Lecture Notes
Classical Mechanics
Ronald Mwesigwa
January 2023
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i
About these notes
The mathematical study of the motion of everyday objects and the forces that affect them is
called classical mechanics. Classical mechanics is often called Newtonian mechanics because
nearly the entire study builds on the work of Isaac Newton. Some mathematical laws and
principles at the core of classical mechanics include the following: Newton’s laws of motion,
Newton’s law of universal gravitation, law of conservation of energy, law of conservation of
momentum, and Bernoulli’s principle.
This document, which is a second version of the handbook of classical mechanics provides
the course instructor and the student of Bachelor of Science with Education with a series of
topics to be covered during the course of study. The document is intended to provide all those
interested in learning classical mechanics with knowledge, skills, and attitudes to enable them
solve problems that involve use of Newton’s laws of motion. Among the chapters included
in this document are linear motion, potential energy, uniform force fields, simple harmonic
motion, forced vibrations, restrained motion, elastic strings and springs, and motion under
a central force.
ii
Acknowledgements
This work was compiled by Ronald Mwesigwa (Senior Lecturer, Mathematics), on behalf of
the Department of Mathematics in the Faculty of Science at Mbarara University of Science
and Technology. However, the information herein has been generated and made available
because of the participation of fellow instructors in teaching other relevant topics to students
at the University. The students have had a tremendous contribution towards the compilation
of these lecture notes for all the time they have attended lectures, after all, they are the main
reason why this is done. The administration, support staff, and instructors from all other
departments in the faculty have been so wonderful, and I guess they will continue to be. Ms.
Sharon Kashemeire, currently secretary to the Vice-chancillor typed and typeset these notes
in Latex. It is indeed a pleasure to thank you all.
The following staff from the Faculty have directly or indirectly contributed to this work
through editing, photocopying, and making useful suggestions: Dr. Simon K. Anguma
(Former Dean, Faculty of Science), Dr. Kasifa Namyalo (Lecturer, Mathematics), Late Dr.
Julius Tumwiine (Associate Professor, Mathematics), Mr. Daniel Baguma (Former Lecturer,
Mathematics), Dr. Eunice A. Olet (Senior Lecturer, Biology), Mr. Michael Byamukama
(Lecturer, Mathematics), Dr. Patrick Mungufeni (Lecturer, Physics), Ms. Mackleane Kyokwebaza (Former Secretary, Faculty of Science), and Ms. Peace M. Mbabazi (Administrator,
Faculty of Science). These notes were read through by Mr. Mohammed Kizito (Assistant
Lecturer, Mathematics), who suggested quite a number of additions in order to have an
improved version of this reading material.
Contents
About these notes
i
Acknowledgements
ii
1 Linear motion
1.1 Introduction . . . . . . . .
1.2 Newton’s laws of motion .
1.3 Inertial frames of reference
1.4 Exercises . . . . . . . . . .
1.5 Answers . . . . . . . . . .
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2 Force fields and acceleration
2.1 Introduction . . . . . . . . . . . . . . . .
2.2 Weight of a particle . . . . . . . . . . . .
2.3 Freely falling bodies . . . . . . . . . . . .
2.4 Potential energy in a uniform force field .
2.5 Exercises . . . . . . . . . . . . . . . . . .
2.6 Answers . . . . . . . . . . . . . . . . . .
3 Conservative forces
3.1 Introduction . . . . . . .
3.2 Conservative force fields
3.3 Potential energy . . . . .
3.4 Conservation of energy .
3.5 Exercises . . . . . . . . .
3.6 Answers . . . . . . . . .
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4 Impulsive forces
4.1 Impulse . . . . . . . . . . . . .
4.2 Torque and angular momentum
4.3 Conservation of momentum . .
4.4 Exercises . . . . . . . . . . . . .
4.5 Answers . . . . . . . . . . . . .
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CONTENTS
iv
5 Work, power and energy
5.1 Work and power . . . .
5.2 Energy . . . . . . . . .
5.3 Exercises . . . . . . . .
5.4 Answers . . . . . . . .
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33
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6 Motion of projectiles
6.1 Introduction . . . . . . . .
6.2 Time of flight and range .
6.3 Range on an inclined plane
6.4 Exercises . . . . . . . . . .
6.5 Answers . . . . . . . . . .
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39
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7 Resisting medium
7.1 Introduction . . .
7.2 Worked examples
7.3 Exercises . . . . .
7.4 Answers . . . . .
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46
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8 Constrained motion
8.1 Introduction . . .
8.2 Friction . . . . .
8.3 Exercises . . . . .
8.4 Answers . . . . .
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53
53
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58
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9 Simple harmonic motion
9.1 Introduction . . . . . . . . . . . .
9.2 Amplitude, period and frequency
9.3 Energy of the harmonic oscillator
9.4 Exercises . . . . . . . . . . . . . .
9.5 Answers . . . . . . . . . . . . . .
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81
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10 The
10.1
10.2
10.3
10.4
2D- and 3D-harmonic
Introduction . . . . . . .
Worked examples . . . .
Exercises . . . . . . . . .
Answers . . . . . . . . .
oscillators
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11 The
11.1
11.2
11.3
11.4
damped harmonic oscillator
Introduction . . . . . . . . . . .
Cases of damping . . . . . . . .
Exercises . . . . . . . . . . . . .
Answers . . . . . . . . . . . . .
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CONTENTS
v
12 Forced vibrations
12.1 Introduction . . .
12.2 Resonance . . . .
12.3 Simple pendulum
12.4 Exercises . . . . .
12.5 Answers . . . . .
13 Elastic strings and
13.1 Introduction . .
13.2 Exercises . . . .
13.3 Answers . . . .
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89
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springs
100
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
14 Momentum and variable-mass problems
14.1 Introduction . . . . . . . . . . . . . . . .
14.2 Mass accretion . . . . . . . . . . . . . .
14.3 The falling raindrop . . . . . . . . . . . .
14.4 The rocket equation . . . . . . . . . . . .
14.5 Exercises . . . . . . . . . . . . . . . . . .
14.6 Answers . . . . . . . . . . . . . . . . . .
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15 Motion under a central force
15.1 Introduction . . . . . . . . . . . . . . . . . .
15.2 Equations of motion for a central force field
15.3 Potential energy in the central force field . .
15.4 Determination of the orbit . . . . . . . . . .
15.5 Exercises . . . . . . . . . . . . . . . . . . . .
15.6 Answers . . . . . . . . . . . . . . . . . . . .
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111
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116
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117
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119
119
123
124
List of Figures
1.1
Particle P as observed from two coordinate systems. . . . . . . . . . . . . . .
5
4.1
Torque of a force. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
5.1
Work done by a force field in moving aa particle from one point to another. .
33
6.1
Range on an inclined plane. . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
8.1
Particle on an inclined plane with friction. . . . . . . . . . . . . . . . . . . .
54
9.1
A horizontal spring-mass system oscillating about O with amplitude A. . . .
60
11.1 Types of damping. . . . . . . . . . . . . . . √
. . . . . . . . . . . . . . . . . .
11.2 Damped motion with maximum amplitude 20 2e−2t . . . . . . . . . . . . . .
82
85
12.1 A swinging pendulum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
13.1 A box on top of horizontal table. . . . . . . . . . . . . . . . . . . . . . . . . 103
13.2 Two springs carrying a mass on top. . . . . . . . . . . . . . . . . . . . . . . 104
13.3 A compressed spring with one end fixed and a teddy bear at the other end. . 105
vi
Chapter 1
Linear motion
1.1
Introduction
Mass of a particle is defined as the quantity of the matter that it contains. A force is a
measure of the push or pull on an object. If m denotes the mass of the particle in some
chosen system of units, and that at a particular instant, the particle has velocity v, then the
vector
p = mv
(1.1)
defines the linear momentum of the particle at the instant. Standard units of mass are
the grammes (g) in the centimeter-gramme-second (c.g.s.) system, kilogramme (kg) in the
metre-kilogramme-second (m.k.s.) system and pound (lb) in the foot-pound-second (f.p.s.)
system. Standard units of force in these systems are the dyne, newton (N) and poundal
(pdl), respectively. A dyne is that force which will give 1 g mass an acceleration of 1 cms−1 .
A newton is the force which will give 1 kg mass an acceleration of 1 ms−1 and a poundal is
that force which will give a 1 lb mass an acceleration of 1 fts−1 .
Here, velocity is defined as the rate of change of displacement, and acceleration as the
rate of change of velocity. This means that, given the position vector or displacement as
a function of time t, you differentiate once to get velocity, whereas a second derivative will
give the acceleration of a particle, all at any time t.
1.2
Newton’s laws of motion
Newton’s three laws of motion talk about the effect of force on motion. These laws are found
to be useful in the subsequent chapters of this manuscript and quite often the reader will be
referred to Newton’s second law of motion, in particular.
1. Law I states that ´’Every particle continues in its state of rest or of uniform velocity in
a straight line unless acted upon by an external force´’. This is sometimes refered to as
the law of inertia, that is, the tendency of a body to resist changes in motion.
1
CHAPTER 1. LINEAR MOTION
2
2. Law II states that ´’The rate of change of momentum is proportional to the exerted or
impressed force and takes place in the direction of the action of force.̈ In other words,
if F is the external force acting on the particle of mass m which as a consequence is
moving with velocity v, then
F=
d
d
(mv) = p.
dt
dt
If m is independent of time, then we write
F = ma,
(1.2)
where a is the acceleration of the particle.
3. Law III states that ´’To every action there is an equal and opposite reaction´’. That is,
if particle 1 acts on particle 2 with a force F12 then particle 2 will react on particle 1
with a force F21 of the same magnitude but in an opposite direction, that is,
F21 = −F12 .
(1.3)
Example 1.1. A particle of mass 2 units moves along the space curve defined by
r = (4t2 − t3 )i − 5tj + (t4 − 2)k.
Find, at time t = 1, (a) the velocity, (b) the momentum, (c) acceleration, and (d) force field.
Solution
(a) Given the position vector r = (4t2 − t3 )i − 5tj + (t4 − 2)k, then v(t) = dr/dt = (8t −
3t2 )i − 5j + (4t3 )k. Now, v(1) = 5i − 5j + 4k
(b) Momentum is given as p = mv, where v = (8t − 3t2 )i − 5j + (4t3 )k, then p(t) =
(16t − 6t2 )i − 10j + (8t3 )k. Now, p(1) = 10i − 10j + 8k
(c) Given the velocity v(t) = (8t−3t2 )i−5j+(4t3 )k, then a(t) = dv/dt = (8−6t)i+(12t2 )k.
Now, a(1) = 2i + 12k
(d) Force is given as F = ma, where a = (8−6t)i+(12t2 )k, then F(t) = (16−12t)i+(24t2 )k.
Now, F(1) = 4i + 24k
Example 1.2. A particle of mass m moves in the xy-plane so that its position vector is
r = a cos ωti + b sin ωtj,
where a, b, and ω are positive constants and a > b. Show that
(a) the particle moves in an ellipse,
CHAPTER 1. LINEAR MOTION
3
(b) the force acting on the particle is directed toward the origin.
Solution
(a) Given the position vector r = a cos ωti + b sin ωtj = xi + yj, then x = a cos ωt and
y = b sin ωt. Now, x2 + y 2 = a2 cos2 ωt + b2 sin2 ωt = 1. Thus,
2 2
x
y
+
= 1,
a
b
which is an equation of the ellipse with major and minor axes of length 2a and 2b,
respectively.
(b) Given the position vector r = a cos ωti + b sin ωtj = xi + yj, then velocity is
v(t) = dr/dt = −aω sin ωti + bω cos ωtj
and acceleration is
a = −ω 2 a cos ωti − ω 2 b sin ωtj = −ω 2 r.
Now, if the particle has a constant mass m, then the force acting on the particle is given
by Newton’s second law as
F = ma = −mω 2 r.
The negative sign shows that the force is always directed in a direction opposite to r,
that is, toward the center.
Example 1.3. A particle of mass 2 units moves in a force field depending on time t given
by
F = 24t2 i + (36t − 16)j − 12tk.
Assuming that at t = 0 the particle is located at 3i − j + 4k and has velocity 6i + 15j − 8k,
find
(a) the velocity, and
(b) the position of the particle, at any time t.
Solution
(a) By Newton’s second law of motion, F = ma or a = F/m. Now, given F = 24t2 i + (36t −
16)j − 12tk, then the acceleration is a = 12t2 i + (18t − 8)j − 6tk. Since a = dv/dt, we
integrate a with respect to t and obtain v = 4t3 i + (9t2 − 8t)j − 3t2 k + c1 , where c1 is
an abitrary constant. Now, at t = 0, v = 6i + 15j − 8k so that c1 = 6i + 15j − 8k.
Therefore, velocity is
v = (4t3 + 6)i + (9t2 − 8t + 15)j − (3t2 + 8)k.
CHAPTER 1. LINEAR MOTION
4
(b) Given that the velocity is v = (4t3 +6)i+(9t2 −8t+15)j−(3t2 +8)k and since v = dr/dt,
we integrate v with respect to t and obtain r = (t4 +6t)i+(3t3 −4t2 +15t)j−(t3 +8t)k+c2 ,
where c2 is an abitrary constant. Now, at t = 0, r = 3i − j + 4k so that c2 = 3i − j + 4k.
Therefore, the position at any time t is
r = (t4 + 6t + 3)i + (3t3 − 4t2 + 15t − 1)j − (t3 + 8t − 4)k.
Example 1.4. A constant force F acting on a particle of mass m changes the velocity from
v1 to v2 in time t. Prove that
m
F = (v2 − v1 )
t
and explain why the result does not hold if the force is variable.
Solution
By Newton’s second law of motion, F = ma or a = F/m. We may write this as
dv
1
= F.
dt
m
Now, if F and m are constants, then by integration with respect to t we have
v=
F
t + c1 .
m
F
At t = 0, v = v1 so that c1 = v1 and thus v = m
t + v1 . Also, at t = t, v = v2 so that
F
v2 = m t + v1 , and therefore
m
F = (v2 − v1 ).
t
Here, the force and mass were taken to be constant, otherwise integration involving F and
m(t) with respect to t would have produced a different result.
1.3
Inertial frames of reference
Newton’s laws are postulated under the assumption that all measurements or observations
are taken with respect to a coordinate system or frame of reference fixed in space or absolutely
at rest. A particle can be at rest or in a uniform motion in a straight line with respect to one
frame of reference and be traveling in a curve and accelerating with respect to another frame
of reference. It can be shown that if Newton’s laws hold in one frame of reference which is
moving at constant velocity relative to it, such frames of reference are called inertial frames
of reference or Newtonian frames of reference. For speeds comparable with the speed of light
299,000 km/s (186,000 miles/s), Newton’s laws of mechanics must be replaced by Einstein’s
laws of relativity.
Consider that two observers O and O′ , fixed relative to two coordinate systems Oxyz
and O′ x′ y ′ z ′ respectively observe the motion of a particle P in space, as shown in Figure 1.1.
It can be shown that to both observers, the particle appears to have the same force acting
CHAPTER 1. LINEAR MOTION
5
Figure 1.1: Particle P as observed from two coordinate systems.
on it if and only if the coordinate systems are moving at constant velocity relative to each
other.
Let the position of the particle in the Oxyz and O′ x′ y ′ z ′ coordinate systems be r and r′
respectively and let the position vector of O′ with respect to O be R = r − r′ . Relative to
observers O and O′ the forces acting on P according to Newton’s law are given by
F=m
d2 r
d2 r ′
′
and
F
=
m
dt2
dt2
The difference in observed force is
F − F′ = m
d2 r′
d2
d2 R
d2 r
′
−
m
=
m
(r
−
r
)
=
m
dt2
dt2
dt2
dt2
and this will be zero if and only if
d2 R
dR
=
0
or
is constant.
dt2
dt
In other words, the coordinate systems are moving at constant velocity relative to each other.
Such coordinate systems are called inertial coordinate systems, and the result is sometimes
called the classical principle of relativity.
CHAPTER 1. LINEAR MOTION
1.4
6
Exercises
1. A particle of mass 2 units moves along the space curve defined by
r = (4t2 − t3 )i + 5tj + (t3 − 4t)k.
Find
(a) the momentum, and
(b) the force acting on it, at t = 1.
2. A particle moving in a force field F has its momentum given at any time t as
p = 3e−t i − 2 cos tj − 3 sin tk.
Find the force F acting on the particle.
3. A particle of mass m moves long the ellipse
r = a cos ωti + b sin ωtj,
under the influence of a force field F. If p is the momentum, prove that
(a) r × p = mabωk
(b) r · p = 21 m(b2 − a2 )ω sin 2ωt
(c) r × F = 0
4. A force of 100 dynes in the direction of positive x-axis acts on a particle of mass 2
grammes for 10 minutes. What velocity does the particle acquire assuming that it
starts from rest?
5. A particle of unit mass moves in a force field given in terms of time t by
F = (6t − 8)i − 60t3 j + (20t3 + 36t2 )k.
Its initial position and velocity are given respectively by r0 = 2i − 3j and v0 = 5i + 4j.
Find, after 2 seconds,
(a) the position,
(b) velocity, and
(c) acceleration, of the particle.
CHAPTER 1. LINEAR MOTION
1.5
7
Answers
1. (a) Momentum, p(t) = (16t − 6t2 )i + 10j + (6t2 − 8)k and p(1) = 10i + 10j − 2k .
(b) Force, F(t) = (16 − 12t)i + 6t)k and F(1) = 4i + 12k .
2. By Newton’s second law of motion, force is given as,
F=
dp
= −3e−t i + 2 sin tj − 3 cos tk.
dt
3. (a) Given the position vector r = a cos ωti + b sin ωtj, then velocity is v(t) = dr/dt =
−aω sin ωti + bω cos ωtj and momentum is p = mv = −maω sin ωti + mbω cos ωtj.
Thus, r × p = mabωk.
(b) r · p = 12 mω(a2 − b2 ) sin 2ωt.
(c) By Newton’s second law of motion, force is given as, F = ma = −mω 2 (a cos ωti +
b sin ωtj), and thus r × p = 0
4. By Newton’s second law of motion, force is given as, F = ma = mdv/dt. So, if
2dv/dt = 100i, then v = 50ti + c (in centimetres per second), and c = 0 since particle
starts at rest. When t = 10 minutes, v = 300i metres per minute.
5. (a) Position is r(t) = (t3 − 4t2 + 5t + 2)i + (4t − 3t5 )j + (t5 + 3t4 − 3)k and r(2) =
4i − 88j + 53k.
(b) Velocity is v(t) = (3t2 − 8t + 5)i + (4 − 15t4 )j + (5t4 + 12t3 )k and v(2) = i − 236j +
176k.
(c) Acceleration is a(t) = (6t−8)i−60t3 )j+(20t3 +36t2 )k and a(2) = 4i−480j+304k.
Chapter 2
Force fields and acceleration
2.1
Introduction
Definition 2.1. A force field is a vector field corresponding with a non-contact force acting
on a particle at various positons in space.
A force field is a force that a particle would feel if it were at that point. Examples of force
fields include the magnetic fields, gravitational fields, electric fields, among others. A force
field which has constant magnitude and direction is called a uniform or constant force field.
If the direction is taken as the negative z-direction and the constant magnitude F0 > 0, then
the force field is given by
F = −F0 k.
If a particle of constant mass m moves in a uniform force field, then its acceleration is
uniform or constant, and the the motion is described as uniformly accelerated motion. Now
by Newton’s second law of motion,
F = ma.
2.2
Weight of a particle
If air resistance is negligible the acceleration near the earth is found to be a constant and
is denoted by g. This acceleration is due to gravity and is sometimes called gravitational
acceleration. The approximate magnitude of g is 9.8 ms−2 as one moves from the equator
toward the poles. Assuming that the surface of the earth is represented by xy-plane, the
force acting on a particle of mass m is given by w = −mgk. This force which is called the
weight of the particle, has magnitude
W = mg.
Since the magnitude of weight is W = mg, we have m = W/g. Thus, by Newton’s second
law of motion we have F = ma. In this equation W and g may vary while m is a constant.
8
CHAPTER 2. FORCE FIELDS AND ACCELERATION
9
In one system of units, force F or weight W is measured in pound weight (1b. wt), length in
feet and time in seconds. In another system, the respective units are kilograms weight (kg
wt), metres (m), and seconds (s).
2.3
Freely falling bodies
If a particle moves so that the only force acting on it is its weight or force due to gravity,
then the particle is often called a freely falling body. If r is the position vector and m is the
mass of the particle, then by Newton’s law of motion we have
m
d2 r
d2 r
=
−mgk
or
= −gk.
dt2
dt2
Note that this equation does not involve the mass m, and thus the motion in a freely falling
particle is independent of mass.
2.4
Potential energy in a uniform force field
The potential of the uniform force field or potential energy of a particle in this force field is
given by
V = F0 (z − z0 ),
where z0 is an arbitrary constant such that when z = z0 , V = 0, we call z = z0 the reference
level. For a constant gravitational field, F0 = mg and potential energy of the particle is
V = mg(z − z0 ).
Definition 2.2. The potential energy of a particle in a constant gravitational field is the
product of the magnitude of its weight and the height above some prescribed reference level.
Note that the potential energy is work done by weight in moving through the distance
z − z0
Example 2.3. A particle of mass m moves along a straight line under the influence of a
constant force of magnitude F . If its initial speed is v0 , find
(a) the speed, and
(b) the distance traveled, after time t.
Solution
CHAPTER 2. FORCE FIELDS AND ACCELERATION
10
(a) Suppose that at time t a particle P is at a distance x from the origin. If i is the unit
vector in the direction of OP and v is the speed at time t, then the velocity is vi. By
Newton’s second law of motion we have
d
d
(mvi) = F i or (mv) = F,
dt
dt
since m is a constant. Integrating and applying the condition v = v0 at t = 0 gives the
speed as
F
v = t + v0 .
m
(b) Here we write
F
dx
= m + v0 .
dt
t
Integrating and applying the condition x = 0 at t = 0 gives the distance as
1 F 2
s=
t + v0 t,
2 m
v=
since m is a constant.
Example 2.4. A particle of mass m moves along a straight line under the influence of a
constant force of magnitude F . If its initial speed is v0 , show that the speed of the particle
at any position x is given by
12
2F
x .
v = v02 +
m
Solution
We know that
a=
dv
dv dx
dv
F
=
=v
= .
dt
dx dt
dx
m
Separation of variables leads to
Z
Z
F
dx.
m
Integrating and applying the initial condition v = v0 at x = 0 gives
vdv =
v2 =
2F
x + v02 .
m
and hence the required result.
Example 2.5. An object of mass m is dropped from a height h above the ground.pProve
that if the air resistance
√ is negligible, then it will reach the ground (a) in time t = 2h/g,
and (b) with speed 2gh.
Solution
CHAPTER 2. FORCE FIELDS AND ACCELERATION
11
(a) Considering the motion of the object along the z-direction, then
d2 z
= −g.
dt2
Integrating and applying the initial condition dz/dt = 0 at t = 0 gives the speed of the
object at any time t as
dz/dt = −gt
Integrating again and applying the condition z = h at t = 0 gives the distance of the
object at any time t as
1
z = h − gt2 .
2
Now, when the object reaches the ground z = 0, and thus 0 = h − 21 gt2 , which gives the
time taken to reach the ground as
s
2h
t=
.
g
(b) The speed
pat any time
√ t is v = gt and so when the particle strikes the ground, this speed
is v = g( 2h/g) = 2gh.
2.5
Exercises
1. An object of mass m is dropped from a height h above the ground. After travelling
through a distance d below the point of projection, a second object os dropped. Prove
that if air resistance is negligible,
when the first object strikes the ground, the second
√
object is at a height of 2 dh − d above the ground.
2. An object of mass m is thrown vertically downward with an initial velocity of magnitude
v0 from a height h above the ground. Assuming that air resistance is negligible, find
the time it will take to reach the ground.
3. A ball which is thrown vertically upward reaches its maximum height of 100 ft and
then returns to the starting point.
(a) With what speed was it thrown?
(b) How long does it take to return?
4. A ball which is thrown vertically upward reaches a particular height h after a time t1
on the way up and a time t2 on the way down. Prove that (a) the initial velocity with
which the ball was thrown has magnitude 21 g(t1 + t2 ), (b) the height h = 12 gt1 t2 , and
(c) the maximum height reached is 18 g(t1 + t2 )2
5. An object of mass m is thrown vertically upward from the earth’s surface with speed
v0 . Prove that it returns to the earth’s surface
CHAPTER 2. FORCE FIELDS AND ACCELERATION
12
(a) with the same speed as the initial speed, and
(b) in a time which is twice that taken to reach the maximum height.
2.6
Answers
1. Considering the motion of the object along the z-direction, then
d2 z
= −g.
dt2
Integrating and applying the initial condition dz/dt = 0 at t = 0 gives the speed of the
object at any time t as
dz
= −gt
dt
Integrating again and applying the condition z = h at t = 0 gives the distance of the
object at any time t as
1
z = h − gt2 .
2
Now, the time t1 taken for the first object to reach a distance d below the cliff is given
by
1
h − d = h − gt21 ,
2
q
which simplifies to t1 = 2d
. The time t2 taken for this object to reach the ground
g
q
. Now we find the distance of the second object
is given by 0 = h − 21 gt22 or t2 = 2h
g
above the ground after time t2 − t1 as
s
s !
1
1
2h
2d
z = h − g(t22 − t21 ) = h − g
−
(2.1)
2
2
g
g
s !
1
2h 2d
hd
= h− g
−
−4
2
g
g
g
√
= h − h − d + 2 hd
√
= 2 hd − d.
2. Considering the motion of the object along the z-direction, then
d2 z
= −g.
dt2
Integrating and applying the initial condition dz/dt = −v0 at t = 0 gives the speed of
the object at any time t as
dz/dt = −gt − v0
CHAPTER 2. FORCE FIELDS AND ACCELERATION
13
Integrating again and applying the condition z = h at t = 0 gives the distance of the
object at any time t as
1
z = − gt2 − v0 t + h.
2
Now, when the object reaches the ground z = 0, and thus 0 = − 21 gt2 − v0 t + h, which
gives the time taken to reach the ground as
p
−2v0 ± 4v02 + 8gh
t=
.
2g
Taking the positive value of t gives
t=
−v0 +
p
v02 + 2gh
.
g
The speed of the object when it reaches the ground is then
p
q
−v0 + v02 + 2gh
v = −gt − v0 = −g
− v0 = v02 + 2gh.
g
3. (a) By Newton’s second law and considering the motion of the object along the zdirection, then
d2 z
= −g.
dt2
Integrating and applying the initial condition dz/dt = v0 at t = 0 gives the speed
of the object at any time t as
dz/dt = −gt + v0
Integrating again and applying the condition z = 0 at t = 0 gives the distance of
the object at any time t as
1
z = − gt2 + v0 t.
2
Now, when the object reaches a height z = 100, the time is t = v0 /g and thus
2
v2
100 = − 21 g vg0 + v0 vg0 = g0 , which gives initial speed as v0 = 80 feet per second.
(b) Now, the time taken to reach maximum height is t = v0 /g = 80/32 = 2.5 seconds,
and time of flight is the sum of time taken to reach the maximum height and the
time taken to travel back to the point of projection. Thus time taken to return
is 5 seconds.
4. (a) By Newton’s second law and considering the motion of the object along the zdirection, then
d2 z
= −g.
dt2
CHAPTER 2. FORCE FIELDS AND ACCELERATION
14
Integrating and applying the initial condition dz/dt = v0 at t = 0 gives the speed
of the object at any time t as
dz/dt = −gt + v0
Integrating again and applying the condition z = 0 at t = 0 gives the distance of
the object at any time t as
1
z = − gt2 + v0 t.
2
Now, when the object reaches a height z = h, the time is t = t1 and t = t2 . So
then h = v0 t − 21 gt21 and h = v0 − 12 gt22 , which gives v0 = 21 g(t1 + t2 ).
(b) Now, substitute for v0 = 21 g(t1 + t2 ) in h = v0 t − 21 gt21 to obtain h = 12 gt1 t2 .
(c) At maximum height, t = v0 /g = 21 g(t1 + t2 )/g = 12 (t1 + t2 ). Now, substitute for
v0 = 12 g(t1 + t2 ) in z = v0 t − 21 gt21 to obtain zmax = 18 (t1 + t2 )2 .
5. (a) Considering the motion of the object along the z-direction and by Newton’s second
law of motion,
d2 z
= −g.
dt2
Integrating and applying the initial condition dz/dt = v0 at t = 0 gives the speed
of the object at any time t as
dz/dt = −gt + v0
Integrating again and applying the condition z = 0 at t = 0 gives the distance of
the object at any time t as
1
z = − gt2 + v0 t.
2
Now, the object returns to the point of projection when z = 0 after the time
t = 2v0 /g. So then the speed is dv/dt = −gt + v0 = −g(2v0 /g) + v0 = −v0 and
this means that the object returns with the same speed.
(b) At maximum height dz/dt = 0 so that t = v0 /g. The time taken to return is
twice time taken to reach the maximum height, that is, tf = 2v0 /g.
Chapter 3
Conservative forces
3.1
Introduction
A conservative force has the property that the work done in moving a particle between two
points is independent of the taken path. Equivalently, if a particle travels in a closed loop,
the net work done by a conservative force is zero. Mathematically, a force field F, defined
everywhere in space (or within a simply-connected volume of space), is called a conservative
force or conservative vector field if it meets any of these three equivalent conditions:
ˆ The curl of F is the zero vector, i.e.,
∇ × F = 0.
(3.1)
ˆ There is zero net work (W) done by the force when moving a particle through a
trajectory that starts and ends in the same place, i.e.,
I
W ≡
F · dr = 0.
(3.2)
C
ˆ The force can be written as the negative gradient of a potential, V , i.e.,
F = −∇V.
3.2
(3.3)
Conservative force fields
Suppose there exists a scalar function V and a force field F such that F = −∇V . Then the
following theorem can be proved.
Theorem 3.1. The total work done in moving the particle along the curve C from P1 to P2
is
Z P2
W =
F · dr = V (P1 ) − V (P2 ).
P1
15
CHAPTER 3. CONSERVATIVE FORCES
16
In such a case the work done is independent of the path joining P1 and P2 . If the work done
by a force field in moving a particle from one point to another point is independent of the
path joining the points, then the force field is said to be conservative.
Proof
If the force acting on a particle is given by F = −∇V , then work done W is obtained as
follows
F = −∇V
Z P2
Z P2
(−∇V ) · dr
F · dr =
W =
P1
P1
Z
P2
=−
P1
dV = −V |PP21
= V (P1 ) − V (P2 )
Theorem 3.2. A force field F is conservative if and only if there exists a continuous differentiable scalar function V such that F = −∇V of equivalently, if and only if ∇ × F = 0
identically.
Proof
If F = −∇V , then
Z
P2
F · dr
Z P2 ∂V
∂V
∂V
=−
i+
j+
k · (dxi + dyj + dzk)
∂x
∂y
∂z
P1
Z P2
∂V
∂V
∂V
dx +
dy +
dz
=−
∂y
∂z
P1 ∂x
Z P2
=−
dV = V (P1 ) − V (P2 )
Work done, W =
P1
P1
Then the integral depends only on the points P1 and P2Rand not on the path joining
them. The function V has to be single valued. Now, suppose C F · dr is independent of the
path C joining any two points, then it can be shown that there exists a scalar function V
such that
F = −∇V ,
that is, F is conservative.
Let F = F1 i + F2 j + F3 k
R
By hypothesis, F · dr is independent of the path C joining any two points which we
may take as (x1 , y1 , z1 ) and (x, y, z) respectively. Then
Z
(x,y,z)
−V (x, y, z) =
Z
(x,y,z)
F · dr =
(x1 ,y1 ,z1 )
(F1 dx + F2 dy + F3 )dz
(x1 ,y1 ,z1 )
CHAPTER 3. CONSERVATIVE FORCES
17
is independent of the path joining (x1 , y1 , z1 ) and (x, y, z). Thus
Z
(x+∆x,y,z)
Z
(x,y,z)
F · dr −
−V (x + ∆x, y, z) + V (x, y, z) =
(x1 ,y1 ,z1 )
Z (x1 ,y1 ,z1 )
F · dr
Z
(x1 ,y1 ,z1 )
(x+∆x,y,z)
F · dr −
=
(x,y,z)
Z (x+∆x,y,z)
F · dr
(x1 ,y1 ,z1 )
F1 dx + F2 dy + F3 dz
=
(x,y,z)
Since the last integral must be independent of the path joining (x, y, z) and (x+∆x, y, z),
we may choose the path to be a straight line joining these points so that dy and dz are zero.
Then
Z (x+∆x,y,z)
V (x + ∆x, y, z) − V (x, y, z))
1
−
=
F1 dx
∆x
∆x (x,y,z)
Taking limits as x → 0, if the limit exists, we obtain
−
∂V
= F1
∂x
Similarly −∂V /∂y = F2 , and −∂V /∂z = F3 . Then
F = F1 i + F2 j + F3 k = −
∂V
∂V
∂V
i−
j−
k = −∇V
∂x
∂y
∂z
Theorem 3.3. A continuous differentiable force field F is conservative if and only if for any
closed non-intersecting curve C,
I
F · dr = 0.
c
Proof
If F is conservative then there exists a scalar differentiable function V such that
F = −∇V
Thus work done is given as
I
F · dr
I
Z
= − ∇V · dr = −
W =
(3.4)
C
C
= V (P1 ) − V (P2 ) = 0
P2
P1
dV
CHAPTER 3. CONSERVATIVE FORCES
18
H
since P1 = P2 on a closed curve. Conversely, if F · dr = 0, then
Z
Z
Z
F · dr =
F · dr +
F · dr
P1 AP2 BP1
P1 AP2
P2 BP1
Z
Z
F · dr −
F · dr
=
P1 AP2
P1 BP2
=0
and thus
Z
Z
F · dr =
P1 AP2
F · dr.
(3.5)
P1 BP2
Therefore the work done is independent of the path.
3.3
Potential energy
Potential energy is the energy possessed by a body by virtue of its position. The scalar
function V such that F = −∇V is called the potential energy. It is also called the scalar
potential or simply the potential of the particle in the conservative force field F. Thus the
total work done from P1 to P2 along the curve C equals the potential energy at P1 minus
the potential energy at P2 , or in symbols
W = V1 − V2 ,
where V1 = V (P1 ) and V2 = V (P2 ). We can write
Z r
V =−
F · dr,
(3.6)
r0
where r0 is some standard position vector and r the final position vector of the particle.
Thus V is a function of the position (x, y, z) of the particle. This means that V = 0 when
r = r0
3.4
Conservation of energy
For a conservative force field we have that the total work done in moving a particle of mass
m from point P1 to P2 along a curve C equals kinetic energy at P2 minus kinetic energy at
P1 . That is,
W = T2 − T1 ,
where T1 = 21 mv12 , T2 = 21 mv22 and v1 and v2 are magnitudes of velocities at P1 and P2
respectively. We also know that the total work done from P1 to P2 along the curve C equals
the potential energy at P1 minus the potential energy at P2 , or
W = V1 − V2 ,
CHAPTER 3. CONSERVATIVE FORCES
19
where V1 and V2 are potential energies at P1 and P2 respectively. Since the work done must
be equal we must have T2 − T1 = V1 − V2 or T1 + V1 = T2 + V2 . This means that
1
1 2
mv1 + V1 = mv22 + V2
2
2
(3.7)
The quantity E = T + V , where T is the kinetic energy and V is the potential, is called
the total energy. This suggests that the total energy at P1 is the same as the total energy
at P2 . This is the principle of conservation of energy. It states that ”In a conservative force
field the total energy is a constant”.
Example 3.4. A force field F defined by
F = (y 2 z 3 − 6xz 2 )i + 2xyz 3 j + (3xy 2 z 2 − 6x2 z)k
(a) Show that F is a conservative force field.
(b) Find the work done by the force field in moving a particle from the point A(−2, 1, 3) to
the point B(1, −2, −1).
Solution
The force F is conservative if and only if curl F = ∇ × F = 0. Now we have
∇×F=
i
j
k
∂
∂x
∂
∂y
∂
∂z
2 2
y 2 z 3 − 6xz 2 2xyz 3 3xy z − 6x2 z
This will reduce to
∇ × F = (6xyz 2 − 6xyz 2 )i + (3y 2 z 2 − 12xz − 3y 2 z 2 + 12xz)j + (2yz 3 − 2yz 3 )k
= 0i + 0j + 0k = 0
Then the force is conservative. Alternatively, the force field F is conservative if and only
if there exists a scalar function V (x, y, z) such that F = −∇V . Then
∂V
∂V
∂V
F=−
i+
j+
k
∂x
∂y
∂z
= (y 2 z 3 − 6xz 2 )i + 2xyz 3 j + (3xy 2 z 2 − 6x2 z)k
Hence if F is conservative we must be able to find V such that
∂V
= 6xz 2 − y 2 z 3 ,
∂x
∂V
= −2xyz 3 ,
∂y
∂V
= 6x2 z − 3xy 2 z 2
∂z
Integrating with respect to x and keeping y and z constant the first equation gives
V = 3x2 z 2 − xy 2 z 3 + g(y, z)
CHAPTER 3. CONSERVATIVE FORCES
20
Differentiating with respect to y and equating to the second equation we obtain
∂V
∂g
= −2xyz 3 +
= −2xyz 3
∂y
∂y
This means that ∂g/∂y = 0 so that g(y) = f (z). So then
V = 3x2 z 2 − xy 2 z 3 + f (z)
Finally differentiating with respect to z and equating to the last equation gives df /dz = 0
or f (z) = c, a constant. Thus,
V = 3x2 z 2 − xy 2 z 3 + c
Example 3.5. A force field F defined by
F = (2xy + z 3 )i + x2 j + 3xz 2 k
(a) Show that F is a conservative force field.
(b) Find the potential energy.
(c) Find the work done in moving an object in this field from (1, −2, 1) to (3, 1, 4).
Solution
(a) A necessary and sufficient condition that a force will be conservative is that ∇ × F = 0.
In this case we have
i
j
k
∂
∂
∂
∇×F=
∂x
∂y
∂z
2xy + z 3 x2 3xz 2
This simplifies to ∇ × F = (0 − 0)i + (3z 2 − 3z 2 j + (2x − 2x)k = 0i + 0j + 0k = 0 and
thus the force is conservative.
(b) The force field F is conservative if and only if there exists a scalar function V (x, y, z)
such that F = −∇V . Then
F=−
∂V
∂V
∂V
i−
j−
k
∂x
∂y
∂t
= (2xy + z 3 i + x2 j + 3xz 2 k
Hence if F is conservative we should be able to find V such that
∂V
= −2xy − z 3 ,
∂x
∂V
= −x2 ,
∂y
∂V
= −3xz 2
∂z
CHAPTER 3. CONSERVATIVE FORCES
21
Integrating with respect to x and keeping y and z constant the first equation gives
V = −x2 y − xz 3 + g(y, z)
Differentiating with respect to y and equating to the second equation we obtain
∂V
∂g
= −x2 +
= −x2
∂y
∂y
This means that ∂g/∂z = 0 and g(z) = f (z), which leads to
V = 3x2 z 2 − xy 2 z 3 + f (z)
Finally differentiating with respect to z and equating to the last equation gives df /dz = 0
or f (z) = c, a constant. Thus,
V = −x2 y − xz 3 + c
(c) Work done is the difference in potentials at the two points. Thus,
W = V (1, −2, 1) − V (3, 1, 4) = 202
Example 3.6. Given a uniform force field F.
(a) Prove that F is conservative.
(b) Find the potential energy corresponding to this field, and
(c) deduce the potential energy of a particle of mass m in a uniform gravitational force field.
Solution
(a) If the force field is due to gravity then
F = −F0 k
But any force field F is conservative if and only if
∇ × F = 0.
Now,
∇×F=
Thus the force field is conservative.
i
j
k
∂
∂x
∂
∂y
∂
∂z
0
0
−F0
=0
CHAPTER 3. CONSERVATIVE FORCES
22
(b) Since F is a conservative force field, there exists a scalar function V such that F =
−∇ × V . Since F = −F0 k, we have
−F0 k = −
Then
∂V
= 0,
∂x
∂V
∂V
∂V
i−
j−
k
∂x
∂y
∂z
∂V
= 0,
∂y
∂V
= F0 ,
∂z
which implies that V = F0 z + c. Now, if V = 0 at z = z0 , then
0 = F0 z0 + c
This implies that c = −F0 z0 and thus
V = F0 (z − z0 )
(c) For uniform gravitational force field F = −mgk. This corresponds to F0 = mg. Thus
the potential energy is
V = mg(z − z0 )
Example 3.7. A particle of mass m moves along the x-axis under the influence of a conservative force field having potential V (x). If the particle is located at positions x1 and x2
at respective times t1 and t2 , prove that if E is the total energy, then
12 Z x2
m
1
p
dx.
t2 − t1 =
2
E − V (x)
x1
Solution
By the principle of conservation of energy, the sum of the kinetic energy and potential energy
equals E, a constant. Thus,
2
m dx
+ V (x) = E
2 dt
This may be written as
dx
dt
2
2
(E − V (x))
m
12
dx
2
=
(E − V (x))
dt
m
m 12
1
dt =
dx
2
E − V (x)
Integration gives the desired result.
=
CHAPTER 3. CONSERVATIVE FORCES
3.5
23
Exercises
1. Determine whether the force field
F = (x2 y − z 3 )i + (3xyz + xz 2 )j + (2x2 yz + yz 4 )k
is conservative
2. A force field F is defined by
F = (y 2 − 2xyz 3 )i + (3 + 2xy − x2 z 3 )j + (6z 3 − 3x2 yz 2 )k
(a) Show that F is a conservative force field.
(b) Find the potential energy V associated with this force.
(c) Find the work done in moving an object in this field from (2, −1, 2) to (−1, 3, −2).
3. Find the work done in moving a particle in the force field
F = 3x2 i + (2xz − y)j + zk
along
(a) the straight line from (0, 0, 0) to (2, 1, 3),
(b) the space curve x = 2t2 , y = t, z = 4t2 − t from t = 0 to t = 1.
4. A particle of mass 4 units moves in the force field defined by
F=−
200
r.
r3
(a) Show that the field is conservative and find the potential energy.
(b) If the particle starts at r = 1 with speed 20, what will be its speed at r = 2?
5. A particle of mass 3 units moves in the xy-plane under the influence of a force field
having potential
V = 12x(3y − 4x).
The particle starts at time t = 0 from rest at the point with position vector 10i − 10j.
(a) Set up the differential equations and conditions describing the motion.
(b) Solve the equations in (a)
(c) Find the position at any time
(d) Find the velocity at any time
6. A force field is given by F = −kr3 r.
CHAPTER 3. CONSERVATIVE FORCES
24
(a) Show that the force field F is conservative.
(b) Write the potential energy of a particle moving in the force field.
(c) If a particle of mass m moves with the velocity v = dx/dt in this field, show that
if E is the constant total energy then
2
1 dr
1
+ kr5 = E.
2 dt
5
3.6
Answers
1. Here, it can be shown that ∇ × F ̸= 0 and therefore the force field is not conservative.
2. (a) F is conservative if ∇ × F = 0, and indeed it is.
(b) Since the force field is conservative, there exists a potential V such that F = −∇V .
It is easy to show, using the method of examples that the potential is
3
V = x2 yz 3 − xy 2 − 3y − z 4
2
(c) Work done is the difference between potential energy at (2, −1, 2) and potential
energy at (−1, 3, −2). At the point (2, −1, 2) we have V = −32−2+3−24 = −55
whereas at (−1, 3, −2) we have V = 24+9−9−24 = 0. Thus, work done = −55−0,
written as 55.
3. (a) Let x = 2t, y = t, z = 3t be the parametric equations for a straight line joining
(0, 0, 0) and (2, 1, 3). Then the position vector is r = 2ti + tj + 3tk, the velocity
is v = 2i + j + 3k and the force was given as F = 3x2 i + (2xz − y)j + zk =
12t2 i + (12t2 − t)j + 3tk. Now, work done is
Z 1
dr
W =
F · dt
dt
0
This gives W = 16.
(b) The position vector is r = 2t2 i+tj+(4t2 −t)k, the velocity is r = 4ti+j+(8t−1)k
and the force was given as F = 3x2 i + (2xz − y)j + zk = 12t2 i + (12t2 − t)j + 3tk.
Now, work done is
Z 1
dr
W =
F · dt
dt
0
and this gives W = 15.8, a different answer from the previous one. This is because
the force field is not conservative, otherwise work done is independent of the path
taken.
CHAPTER 3. CONSERVATIVE FORCES
25
4. F is conservative if ∇ × F = 0, and indeed it is. In this case
3
200
F = − 3 r = −200(x2 + y 2 + z 2 )− 2 (xi + yj + zk)
r
The potential energy V is given by
Z
V = − F · dr
Z r
−200 3 · dr
= −
r
Z
1
dr
= 200
r2
200
= −
r
where the constant of integration is taken to be zero.
(3.8)
5. (a) Given that V = 12x(3y − 4x) = 36xy − 48x2 , then the force F = −∇V =
(96x − 36y)i − 36xj. By Newton’s second law we have
d2 x
= 32x − 12y
(3.9)
dt2
d2 y
= −12x
(3.10)
dt2
These equations are the required differential equations, and the initial conditions
are x = 10, y = −10, dx/dt = dy/dt = 0 at t = 0.
(b) These equations can be solved to get y = −2(exp (6t) + exp (−6t)) − 6 cos 2t and
x = 6(exp (6t) + exp (−6t)) − 2 cos 2t
(c) The position at any time t is r = xi + yj = (6(exp (6t) + exp (−6t)) − 2 cos 2t)i +
(−2(exp (6t) + exp (6t)) − 6 cos 2t)j = (12 cosh 6t − 2 cos 2t)i − (4 cosh 6t + 6 cos 2t)j
(d) Velocity is v = dr/dt so that
v = (72 sinh 6t − 4 sin 2t)i − (24 sinh 6t − 12 sin 2t)j
6. (a) F is conservative if ∇ × F = 0, and indeed it is. In this case
F = −kr3 r = −k(x2 + y 2 + z 2 )3 (xi + yj + zk)
The potential energy V is given by
Z
V =−
F · dr
Z
= − −kr3 r · dr
Z
= k r4 dr
=
k 5
r +c
5
CHAPTER 3. CONSERVATIVE FORCES
26
where the constant of integration, c may be taken to be zero.
(b) By the principle of conservation of energy, the sum of the kinetic energy and
potential energy is a constant, and so
2
1
1 dr
+ kr5 = E,
2 dt
5
where E is the total energy.
Chapter 4
Impulsive forces
4.1
Impulse
The impulse of a force acting on a particle in any interval of time is defined to be the
momentum change produced. It is the product of force and time during which the force
acts. Thus, if a particle of constant mass m has a velocity changed from v1 to v2 in a time
t by a force F acting, then the impulse I is given by
I = mv2 − mv1
Z t2 Z t2
dv
=m
dt =
Fdt,
dt
t1
t1
since F = m(dv/dt). Thus, the impulse of the force F is the time-integral of the force. Now
suppose that F grows very large, the interval t2 − t1 will be very small but in such a way as
to ensure that the time-integral I stays finite. Such forces are called impulsive forces.
Theorem 4.1. The impulse is equal to the change in momentum, that is,
Z t2
Fdt = mv2 − mv1 = p2 − p1 .
t1
Proof: By Newton’s second law of motion we have,
F=m
so that
Z
t2
Z
t2
Fdt =
t1
t1
=
dv
d2 r
=m
2
dt
dt
dv
m dt =
dt
m [v]tt21
Z
t2
t1
d
(mv)dt
dt
= mv(t2 ) − mv(t1 )
= mv2 − mv1 = p2 − p1
27
CHAPTER 4. IMPULSIVE FORCES
4.2
28
Torque and angular momentum
Torque is the measure of the force that can cause an object to rotate about an axis. In other
words, torque is the twisting force that tends to cause rotation. The point where the particle
rotates is known as the axis of rotation. If a particle with position vector r moves in a force
field F, as shown in Figure 4.1, the torque or moment of the force F about the origin O, is
defined by
q = r × F.
(4.1)
The magnitude of q is a measure of the turning effect produced on the particle by the force.
Figure 4.1: Torque of a force.
Theorem 4.2. The torque acting on the particle equals the time rate of change in the angular
momentum.
Proof: The moment of force or torque about the origin O is
q=r×F=r×
d
(mv).
dt
The angular momentum or the moment of momentum about O is
Ω = m(r × v) = r × (mv)
dΩ
d
=
(r × (mv))
dt
dt
dr
d
=
× mv + r × (mv)
dt
dt
d
= mv × v + r × (mv)
dt
But since v × v = 0, we have
dΩ
d
= r × (mv) = r × F = q,
dt
dt
which is the rquired result.
(4.2)
CHAPTER 4. IMPULSIVE FORCES
If we let q = 0 in the equation
dΩ
dt
29
= q then
d
dΩ
= (mr × v) = q = 0
dt
dt
or mr × v = constant. This proves the principle of conservation of angular momentum,
which states that ´’If the net external torque acting on a particle is zero, then the angular
momentum will remain unchanged´’.
4.3
Conservation of momentum
In Newton’s second law of motion, if the external force is zero, i.e., F = 0 then
d
(mv) = 0
dt
or mv = c, a constant. This proves the principle of conservation of momentum, which
states that ´’If the net external force acting on a particle is zero, its momentum will remain
unchanged´’.
Example 4.3. A particle of unit mass moves in a force field given by
F = (3t2 − 4t)i + (12t − 6)j + (6t − 12t2 )k,
where t is time.
(a) Find the change in momentum of the particle from time t = 1 to t = 2.
(b) If the velocity at t = 1 is 4i + −5j + 10k, what is the velocity at t = 2.
Solution
(a) Impulse or change in momentum is given as
Z t2
Z 2
Fdt =
((3t2 − 4t)i + (12t − 6)j + (6t − 12t2 )k)dt
t1
1
3
2
2
3
2
= (t − 2t)i + (6t − 6t)j + (3t − 4t )k
1
= (0i + 12j − 20k) − (−i + 0j + k)
= i + 12j − 19k
(b) By Newton’s second law of motion, since m = 1,
dv
= (3t2 − 4t)i + (12t − 6)j + (6t − 12t2 )k,
dt
CHAPTER 4. IMPULSIVE FORCES
30
Then
v = (t3 − 2t2 )i + (6t2 − 6t)j + (3t2 − 4t3 )k + c.
At t = 1, v = 4i − 5j + 10k so that 4i − 5j + 10k = −i + 0j − k + c giving c =
5i − 5j + 11k. Thus, v(t) = (t3 − 2t2 + 5)i + (6t2 − 6t − 5)j − (3t2 − 4t3 + 11)k and so
v(2) = (8 − 8 + 5)i + (24 − 12 − 5)j + (12 − 32 + 11)k = 5i + 7j − 9k.
Example 4.4. A particle of mass 2 moves in a force field depending on t given by
F = 24t2 i + (36t − 16)j − 12tk.
Determine
(a) the torque, and
(b) the angular momentum, about the origin at any time t assuming that at t = 0, the
particle is located at r0 = 3i − j + 4k and has velocity v0 = 6i + 15j − 8k.
Solution
(a) Given that F = 24t2 i+(36t−16)j−12tk, then since particle is of mass 2, the acceleration
is a = 12t2 i + (18t − 8)j − 6tk. By integrating and applying the initial conditions, we
obtain v = (4t3 + 6)i + (9t2 − 8t + 15)j − (3t2 + 8)k and r = (t4 + 6t + 3)i + (3t3 − 4t2 +
15t + 1)j − (t3 + 8t − 4)k. Then the torque is q = r × F = (32t3 + 108t2 − 200t + 64)i +
(12t5 + 19t3 − 168t2 − 30t)j − (36t5 − 80t4 + 360t3 − 240t2 − 12t − 148)k
(b) Angular momentum, Ω = r × mv = m(r × v) = (8t4 + 36t3 − 130t2 + 64t − 104)i − (2t6 +
48t4 − 56t3 − 18t2 − 96)j − (6t6 − 16t5 + 90t4 − 80t3 − 6t2 + 48t − 102)k
Example 4.5. A particle of mass m moves along a space curve defined by
r = a cos ωti + b sin ωtj.
Find
(a) the torque, and
(b) the angular momentum, about the origin.
Solution
(a) Given the position vector r = a cos ωti + b sin ωtj = xi + yj, then velocity is
v(t) = dr/dt = −aω sin ωti + bω cos ωtj
and acceleration is
a = −ω 2 a cos ωti − ω 2 b sin ωtj.
By Newton’s second law of motion, F = −mω 2 a cos ωti − mω 2 b sin ωtj. Then the torque
is q = r × F = 0i + 0j − 0k.
(b) Angular momentum, Ω = r × mv = m(r × v) = mabωk
CHAPTER 4. IMPULSIVE FORCES
4.4
31
Exercises
1. A particle of unit mass moves in a force field given by
F = (3t2 − 4t)i + (12t − 6)j + (6t − 12t2 )k,
where t is time. Assuming that at t = 0 the particle is located at the origin, find
(a) the torque, and
(b) the angular momentum about the origin at the time t = 2.
2. A particle moves in a force given by F = ϕ(r)r. Prove that the angular momentum of
the particle about the origin is constant.
3. Find the impulse developed by a force given by
F = 4ti + (6t2 − 2)j + 12k
from t = 0 to t = 2.
4. At t = 0 a particle of unit mass m is at rest at the origin. If it is acted upon by a force
F = 100te−2t i, find
(a) the change in momentum of the particle in going from time t = 1 to t = 2,
(b) the velocity after a long time has elapsed.
5. A particle moves in a force field given by F = r2 r where r is the position vector of the
particle. Prove that the angular momentum of the particle is conserved.
6. The angular momentum of a particle is given as a function of time t by
Ω = 6t2 i − (2t + 1)j + (12t3 − 8t2 )k.
Find the torque at the time t = 1.
4.5
Answers
1. (a) Given that F = (3t2 − 4t)i + (12t − 6)j + (6t − 12t2 )k, then since particle is of unit
mass, the acceleration is a = (3t2 − 4t)i + (12t − 6)j + (6t − 12t2 )k. By integrating
twice and applying the initial conditions, we obtain
4
2 3
t
− t + 5 i + (2t3 − 3t2 + 5t)j + (t3 − t4 + 11t)k.
r=
4
3
Now, r(2) =
26
i
3
+ 6j + 14k and F(2) = 4i + 18j − 36k
CHAPTER 4. IMPULSIVE FORCES
32
(b) The torque is given as q = r × F and thus q(t) = (−12t5+ 30t4 + 48t3 − 162t2 +
66t)i + (2t3 − 3t2 + 5t)j + −3t5 + 15
t4 7t3 + 40t2 − 30t k and q(2) = −36i +
2
368j + 180k. Also, since q =
dΩ/dt then Ω(t) = (−2t6 + 6t5 +
12t4 − 54t3 +
5 6
t + 93
t4 − 74
t3 j + − 21 t6 + 32 t5 + 74 t4 + 40
t3 − 15t2 k and Ω(2) =
33t2 )i + − 12
4
3
3
232
−44i + 148j + 3 k.
2. The force is given by F = ϕ(r)r and the torque is q = r × F = 0. Now, since
q = dΩ/dt = 0, then Ω is a constant vector.
3. Impulse,
Z
2
F(t)dt = 8i + 12j + 24k.
I=
0
4. (a) Impulse or change in momentum is
I = (75e−2 − 125e−4 )i.
(b) The velocity after a long time is 25i.
5. The force is given by F = r2 r and the torque is q = r × F = 0. Now, since q =
dΩ/dt = 0, then Ω is a constant vestor.
6. Given that Ω = 6t2 i − (2t + 1)j + (12t3 − 8t2 )k and q = dΩ/dt, then q(t) = 12ti − 2j +
(36t2 − 16t)k. Thus, q(1) = 12i − 2j + 20k.
Chapter 5
Work, power and energy
5.1
Work and power
If a force F shown in Figure 5.1 acts on a particle to give it a displacement dr then the
element of the work denoted by dW done by the force on the particle is defined as
dW = F · dr
(5.1)
Figure 5.1: Work done by a force field in moving aa particle from one point to another.
The total work done by a force field F in moving a particle from point P1 to point P2
along the curve C is given by the integral
Z
Z P2
Z r2
W =
F · dr =
F · dr =
F · dr
(5.2)
C
P1
r1
where r1 and r2 are the position vectors of P1 and P2 respectively. If F and r are functions
of time, then
dr
dr = dt = vdt
dt
33
CHAPTER 5. WORK, POWER AND ENERGY
where the velocity v = dr/dt and therefore the work done is given as
Z t2
F · dvdt
W =
34
(5.3)
t1
where t1 and t2 are the times when the particle is at P1 and P2 respectively.
The time rate of doing work on a particle is often called instantaneous power or briefly
the power applied to the particle. If W is work done and P is the power, then
P =
dW
.
dt
(5.4)
If F is the force acting on a particle and v is the velocity of the particle, then
P =F·
dr
= F · v,
dt
since v = dr/dt.
5.2
Energy
Energy is the ability to do work. In this section we shall look at one type of mechanical
energy called kinetic energy, since potential energy has already been considered. This is the
energy possessed by a body by virtue of its motion. Suppose that a particle has a constant
mass m and that at times t1 and t2 it is located at P1 and P2 and moving with velocities v1
and v2 , respectively. It can be proved that the total work done in moving the particle along
a curve from P1 to P2 (in Figure 5.1) is given as
Z
1
F · dr = m(v22 − v12 )
W =
2
C
The quantity T = 21 mv2 = 12 mv 2 is called the kinetic energy of the particle. This result
is equivalent to the statement ”Total work done from P1 to P2 along C equals kinetic energy
at P2 minus kinetic energy at P1 ”, or in symbols
1
1
W = mv22 − mv12
2
2
Example 5.1. A particle of mass m moves under the influence of the force given by
F = a(sin ωti + cos ωtj).
If the particle is initially at rest at the origin, prove that
(a) the work done W on the particle up to time t is given by
a2
(1 − cos ωt),
mω 2
(5.5)
CHAPTER 5. WORK, POWER AND ENERGY
35
(b) the instatntaneous power applied to the particle is given by
a2
sin ωt.
mω
Solution
By Newton’s second law of motion, we have F = ma = a(sin ωti + cos ωtj), and thus
a
a= m
(sin ωti + cos ωtj). Since a = d2 r/dt2 , we integrate a with respect to t and obtain
a
dr
=
(− cos ωti + sin ωtj) + c1 ,
dt
mω
where c1 is a constant of integration. At t = 0, dr/dt = 0 and thus −(a/mω)i + c1 = 0, or
c1 = (a/mω)i. Therefore,
a
dr
=
[(1 − cos ωt)i + sin ωtj].
dt
mω
Now, the work done from time t = 0 to any time t is given as
Z t
dr
F · dt
W =
dt
Z0 t
a
a(sin ωti + cos ωtj) ·
((1 − cos ωt)i + sin ωtj)dt
=
mω
0
Z t
a2
(sin ωt − sin ωt cos ωt + cos ωt sin ωt)dt
=
mω 0
Z t
a2
=
sin ωtdt
mω 0
t
−a2 cos ωt
=
mω
ω 0
2
a
(1 − cos ωt)
=
mω 2
Example 5.2. Find the instantaneous power applied to a particle of mass 5 units moving
along a space curve whose position vector is given as a function of time t by
r = (2t3 + t)i + (3t4 − t2 + 8)j − 12t2 k.
Solution
The position vector, which is also refered to as the displacement is given as
r = (2t3 + t)i + (3t4 − t2 + 8)j − 12t2 k.
The velocity, v is the rate of change of displacement, or v = dr/dt = (6t2 +1)i+(12t3 −2t)j−
24tk and the acceleration a is the rate of change of velocity, or a = dv/dt = 12ti+(36t2 −2)j−
24k. By Newton’s second law of motion, the force F is given by F = ma = 60ti + (180t2 −
10)j − 120k. Then power P at any time t is given by P = F · v = 2160t5 − 120t3 + 2960t.
CHAPTER 5. WORK, POWER AND ENERGY
36
Example 5.3. A particle of mass m moves in the xy-plane so that its position vector is
r = a cos ωti + b sin ωtj,
where a, b and ω are positive constants and a > b. Find
(a) the kinetic energy at x = a and y = b,
(b) the work done by the force field in moving the particle from x = a to y = b.
Solution
(a) Given the position vector r = a cos ωti + b sin ωtj, then velocity is v(t) = dr/dt =
−aω sin ωti + bω cos ωtj and kinetic energy is given as
1
1
K.E. = mv 2 = mω 2 (a2 sin2 ωt + b2 cos2 ωt).
2
2
Since the path is an ellipse with lengths of major and minor axes given as 2a and
2b respectively, then at x = a, cos ωt = 1 and sin ωt = 0, and kinetic energy, K.E. =
1
mω 2 b2 . Again, at y = b, cos ωt = 0 and sin ωt = 1, and kinetic energy, K.E. = 12 mω 2 a2 .
2
(b) Work done in moving the particle from point A(a, 0) to point B(0, b) is the difference in
kinetic energy at the two points. Thus,
1
W = mω 2 (a2 − b2 ).
2
5.3
Exercises
1. A particle is moved by a force F = 20i − 30j + 15tk along a straight line from point A
to point B with position vectors 2i + 7j − 3k and 5i − 3j − 6k respectively. Find the
work done.
2. Find the kinetic energy of a particle of mass 20 units moving with velocity 3i − 5j + 4k.
3. Due to a force field F, a particle of mass 4 units moves along the space curve r =
(3t2 − 2t)i + t3 j − t4 k. Find the work done by the field in moving the particle from the
point where t = 1 to the point where t = 2.
4. A particle moves with velocity v = 5i − 3j + 6k under the influence of a constant force
F = 20i + 10j + 15k. What is the instantaneous power applied to the particle?
5. A body of mass m, traveling in a straight line, is supplied with constant power P and
is subjected to a resistance kmv 2 , where v is its speed and k is a constant.
CHAPTER 5. WORK, POWER AND ENERGY
37
(a) Prove that the speed of the body cannot exceed a certain value and that, if it
starts from rest it acquires half the maximum speed after traveling a distance
8
1
ln
.
3k
7
(b) If the power is then cut off and an additional retarding force of constant value F
is imposed, find the subsequent time which elapses before the body comes to rest.
5.4
Answers
1. Work done is given as
W = F · r,
where r = (5i − 3j − 6k) − (2i + 7j + 3k = 3i − 10j − 3k. Thus, work done is
W = (20i − 30j + 15k) · (3i − 10j − 3k) = 315
2. Kinetic Energy = 21 mv 2 and v 2 = v · v = (3i + 5j + 4k) · (3i + 5j + 4k) = 50. So kinetic
energy = 12 × 20 × 50 = 500.
3. Work done, W is given as
Z
2
F·
W =
1
dr
dt.
dt
2
Given the posion vector r = (3t − 2t)i + t3 j − t4 k, then dr/dt = (6t − 2)i + 3t2 j − 4t3 k
and acceleration is a = 6i + 6j − 12t2 k. By Newton’s second law of motion we have
F = ma = 24i + 24j − 48t2 k. Thus, work done is 2454.
4. The instantaneous power is P = 20i + 10j + 15k · 5i − 3j + 6k = 160
5. By Newton’s second law of motion we have
m
d2 x
P
= − mkv 2 .
2
dt
v
This may also be written as
mv
dv
P
mv 2
= − mkv 2 or dx =
dv,
dx
v
P − mkv 3
by separation of variables. Integrating and then simplifying gives
P − kmv 3 = exp (−3k(x − c))
CHAPTER 5. WORK, POWER AND ENERGY
38
where c is a constant of integration. Since exponential function is positive we have
P − kmv 3 > 0 or P > kmv 3 . Thus the velocity cannot exceed
1
P 3
km
In other words, this is the maximum speed. Now, we have P −kmv 3 = exp (−3k(x − c)).
1
ln P . Thus,
At t = 0, v = 0, and x = 0 so that P = exp (3kc) or c = 3k
P − kmv 3 = P exp (−3kx).
When the speed is half the maximum it means that we substitute for v =
P
v 3 = 8km
and this gives
8
1
ln
.
x=
3k
7
13
1 P
2 km
or
Again by Newton’s second law of motion we have
m
dv
= −mkv 2 − F,
dt
where F is the magnitude of the retarding force imposed. Then,
dt = −
m
dv.
F + mkv 2
Integrating and then simplifying gives
21 !
12
kF
m
arctan
+c
t=−
kF
m
1
P 3
where c is a constant of integration. Now, at t = 0, v = 12 km
. When the particle
comes to rest we have v = 0 so time taken is
1 1 !
1 km 2 P 3
m
arctan
t=
kF
2 F
km
Chapter 6
Motion of projectiles
6.1
Introduction
An object fired from a gun or dropped from a moving airplane is often called a projectile.
Suppose that a particle of mass m is projected with a velocity v0 at an angle α with the
horizontal. The problem is to find the time of flight of the particle and the range on a
horizontal plane. The following assumptions have been made:
1. The gravitational acceleration g is constant.
2. The forces due to air resistance can be neglected.
3. There is no effect due to the rotation of the earth.
6.2
Time of flight and range
Let Fy be the component of the force acting along y-axis and Fz be the component along
the z-axis. Then by Newton’s second law of motion we have
Fy = m
d2 y
d2 z
=
0
and
F
=
m
= −mg
z
dt2
dt2
Now, along the direction of the y-axis, we have
m
d2 y
=0
dt2
d2 y
=0
dt2
y = c1 t + c2
At t = 0, y = 0 and so c2 = 0. Thus,
y = c1 t.
39
CHAPTER 6. MOTION OF PROJECTILES
Again, at t = 0,
dy
dt
40
= v0 cos α and so c1 = v0 cos α. Thus,
y = (v0 cos α)t.
(6.1)
Similary, taking components along the z-axis leads to
d2 z
m 2 = −mg
dt
dz
= −gt + c3
dt
1
z = − gt2 + c3 t + c4
2
At t = 0, z = 0 and so c4 = 0. Thus,
1
z = − gt2 + c3 t.
2
Again, at t = 0,
dz
dt
= v0 sin α and so c3 = v0 sin α. Thus,
1
z = − gt2 + (v0 sin α)t.
(6.2)
2
The equation of the path traveled is obtained by eliminating t from (6.1) and (6.2). Now,
from (6.1) we have
y
t=
v0 cos α
Substituting this value in equation (6.2) we have
1
y
y
2
z = − g(
) + (v0 sin α)
2 v0 cos α
v0 cos α
(6.3)
2
gy
= y tan α − 2
2v0 cos2 α
Since α, v0 , and g are constants equation (6.3) represents a parabola. The time taken for
the projectile to attain maximum height is the time it takes for its velocity to become zero
for the first time after projection. This is obtained by setting dz/dt = 0. Thus,
dz
= −gt + v0 sin α = 0
dt
v0 sin α
t=
g
The time of flight is the time it takes for the object to reach the horizontal axis again.
This is obtained by setting z = 0 in case projectile lands at a point which is at the same
level with the point of projection. If tf is the time of flight then
1
z = − gt2 + (v0 sin α)t = 0
2
2v0 sin α
t=
g
CHAPTER 6. MOTION OF PROJECTILES
41
Then the distance travelled along the plane, called the range R is given by
R = (v0 cos α)tf
2v0 sin α
g
2
2v cos α sin α
= 0
g
2
v sin 2α
= 0
g
= (v0 cos α) ·
The maximum range occurs when sin 2α is maximum, i.e., sin 2α = 1 or α = π/4.
6.3
Range on an inclined plane
To find the range of a projectile on an inclined plane we find the point of intersection of the
inclined plane and the parabolic path of flight, Figure 6.1. Since (y1 , z1 ) lies on the path of
the projectile we have
gy12
z1 = y1 tan α − 2
.
2v0 cos2 α
Figure 6.1: Range on an inclined plane.
We also have z1 = y1 tan β, where β is the angle between the inclined plane and the
horizontal plane. Then we have
y1 tan β = y1 tan α −
gy12
.
2v02 cos2 α
CHAPTER 6. MOTION OF PROJECTILES
42
This gives
y1 =
2v02 cos2 α
(tan α − tan β).
g
The range on the incline plane OP is
2v 2 cos2 α
y1
= 0
(tan α − tan β)
cos β
g cos β
2v02 cos2 α sin α cos β − cos α sin β
=
g cos β
cos α cos β
2
2v sin(α − β) cos α
= 0
g cos2 β
OP =
For maximum range we use the trigonometric identity sin A cos B =
sin(A − B)) to write
v02
OP =
(sin(2α − β) − sin β).
g cos2 β
1
(sin(A
2
+ B) +
Thus OP is maximum when sin(2α − β) is maximum, that is, when sin(2α − β) = 1 and this
menas that 2α − β = π/2 or α = π4 + β2 . The value of maximum range OP is then
OPmax
v02
(1 − sin β)
=
g cos2 β
v02 (1 − sin β)
=
g(1 − sin2 β)
v02
=
g(1 + sin β)
Example 6.1. A projectile is launched with a muzzle velocity of 980 m/s at an angle of 60o
with horizontal and lands on the same plane. Find
(a) the maximum height reached,
(b) the total time of flight,
(c) the range,
(d) the speed after 60 seconds,
(e) the speed at a height 24500 metres.
Solution
The equations governing the motion of projectiles are y = v0 cos αt along the horizontal
direction, and z = v0 sin αt − 12 gt2 along the vertical direction.
CHAPTER 6. MOTION OF PROJECTILES
43
(a) The distances travelled √
along the y− and z−axes are y = 980 cos 60o t = 490t and z =
1
980 sin 60o t − 2 t2 = 490 3t − 12 gt2 , respectively. The maximum height is reached when
√
√
√
−2
g
=
9.8ms
,
we
obtain
t
=
490
3/9.8
=
50
3. The
dz/dt = 0 or t = 490 3/g. Taking
√
√
√
√
maximum height is zmax = 490 3t − 21 t2 = 490 3(50 3) − 12 (50 3)2 = 36750 metres.
(b) In this case, total time of flight is twice the time to reach the maximum height, that is,
√
2 × 50 3 = 173.2 seconds.
(c) The range R is the horizontal distance covered during the time of flight, that is R =
490 × 173.2 = 84870 metres.
(d) The speed of the projectile along the y− and z−directions
are vy = dy/dt = 980 cos 60o =
√
490 and vz = dz/dt = 980 sin √
60o − gt = 490 3 − 9.8t, respectively. So then, after 60
seconds vy = 490 and
√ vz = 490 3 − 9.8 × 60 = 260.7. Therefore the speed of the particle
after 60 seconds is 4902 + 260.72 = 555.0 metres per second.
(e) Substitute for z = 24500 in z = v0 sin αt − 12 gt2 to obtain t1 = 36.6, when the particle is
going up, and t2 = 136.6, when the particle is going down. We shall use t1 = 36.6 since
both values of t will lead to the same
answer. Use vy = dy/dt = 980 cos 60o = 490 and
√
vz = dz/dt = 980 sin 60o − gt = 490 3 − 9.8t with t = 36.6 to obtain the speeds in both
directions as vy = 490
√ and vz = 490, and therefore the speed of the particle at a height
of 24500 metres is 4902 + 4902 = 692.9 metres per second.
Example 6.2. A projectile having horizontal range R reaches maximum height H. Prove
that it must have been launched with
(a) an initial speed equal to
r
g(R2 + 16H 2
8H
(b) at an angle with the horizontal given by
4H
arcsin √
R2 + 16H 2
Solution
The equations governing the motion of projectiles are y = v0 cos αt along the horizontal
direction, and z = v0 sin αt − 21 gt2 along the vertical direction. The maximum height is
α
attained when dz/dt = 0, that is, v0 sin α − gt = 0 or t = v0 sin
. So, if H is the maximum
g
height, then
2
v0 sin α 1 v0 sin α
v 2 (1 − cos 2α)
H = (v0 sin α)
− g
= 0
.
g
2
g
4g
Now, the time of flight is 2v0 sin α/g, so the range is
R = (v0 cos α)
2v0 sin α
v 2 sin 2α
= 0
.
g
g
CHAPTER 6. MOTION OF PROJECTILES
44
This then means that cos 2α = 1 − 4gH/v02 and sin 2α = gR/v0 . Now, use cos2 2α +
sin 2α = 1 to eliminate α from these two equations and get
r
g(R2 + 16H 2
.
v0 =
8H
p
Then substitute for v0 = g(R2 + 16H 2 )/8H into H = v02 sin2 α/2g to obtain
4H
α = arcsin √
.
R2 + 16H 2
2
6.4
Exercises
1. Find the maximum range possible and height reached for a projectile fired from a
cannon having muzzle velocity of 1,960 ms−1 .
2. A cannon has its maximum range given by Rmaxp
. Prove that (a) the height reached in
1
such case is 4 Rmax and (b) the time of flight is Rmax /2g.
3. A projectile is launched at an angle α from a cliff of height h above sea-level. If it falls
into the sea at a distance d from the base of the cliff, prove that its maximum height
above sea-level is
d2 tan2 α
.
h+
4(h + d tan α)
6.5
Answers
1. From the previous examples, range is given as
R=
v02 sin 2α
.
g
The maximum value of R is attained when sin 2α = 1 and thus Rmax = v02 /g = 392
km. The maximum height reached is Hmax = v02 /4g = 98 km.
2. (a) We know that the equations governing the motion of projectiles are y = v0 cos αt
along the horizontal direction, and z = v0 sin αt− 21 gt2 along the vertical direction.
The maximum height is attained when dz/dt = 0, that is, v0 sin α − gt = 0 or
v 2 sin 2α
α
. So, if R is the range then R = 0 g , which is maximum when
t = v0 sin
g
sin 2α = 1, that is, α = π/4 and Rmax = v02 /g. The maximum height zmax may
be obtained by substituting for α = π/4 and v02 = gRmax in z = v0 sin αt − 21 gt2
to get
1
zmax = Rmax .
4
CHAPTER 6. MOTION OF PROJECTILES
45
(b) First note that the time of flight here is twice the time to attain maximum height,
2
and
p the substitute for α = π/4 and v0 = gRmax in tf = 2v0 sin α/g to get tf =
Rmax /2g.
3. Remember that the equations governing the motion of projectiles are y = v0 cos αt
along the horizontal direction, and z = v0 sin αt − 21 gt2 along the vertical direction.
Now, when the projectile reaches sea-level, then z = −h so that −h = v0 sin αt − 21 gt2 ,
and this gives
p
2v0 sin α ± 4v02 sin2 α + 8gh
.
t=
2g
p
Taking the positive value of time gives t = v0 sin α + v02 sin2 α + 2gh /g. The
horizontal distance covered during this time is then
q
v0 cos α 2
2
d=
v0 sin α + v0 sin α + 2gh .
g
This may be simplified as follows:
gd
v0 cos α
q
gd
= v0 sin α + v02 sin2 α + 2gh
v0 cos α
q
− v0 sin α = v02 sin2 α + 2gh
g 2 d2
− 2gd tan α + v02 sin2 α = v02 sin2 α + 2gh
v02 cos2 α
gd2
− 2d tan α = 2h
v02 cos2 α
and thus
v02
gd2
=
2(h + d tan α) cos2 α
The maximum height is attained when dz/dt = 0, that is, v0 sin α−gt = 0 or t =
So, if zmax is the maximum height, then
zmax =
v0 sin α
.
g
v02 sin2 α
.
2g
Now, substituting for v0 in the equation for zmax gives the maximum height above point
of projection as
zmax =
gd2
sin2 α
d2 tan2 α
=
2(h + d tan α) cos2 α 2g
4(h + d tan α)
and adding h gives the maximum height above sea-level as required.
Chapter 7
Resisting medium
7.1
Introduction
Unless a particle is in a vacuum, it is acted upon by external forces. An important class of
forces are those which tend to oppose the motion of a particle. These are called resisting,
damping or dissipative forces. The corresponding medium is so called respectively. If the
resisting force is R then the motion of a particle of mass m in an otherwise (gravitational)
force field is given by the equation
m
7.2
d2 r
= mgk − R
dt2
Worked examples
Example 7.1. At time t = 0 a parachutist having weight of magnitude mg is located at
z = 0 and is traveling vertically downward with speed v0 . If the force or resistance acting on
the parachute is proportional to its instantaneous speed, find (a) the speed, (b) the distance
traveled, and (c) the acceleration, at any time t > 0.
Solution
Assume that the parachutist is located at distance z from the origin O. If k is a unit vector
in the vertically downward direction, then the weight is mgk while the force of air resistance
is −βvk where β is a constant of proportionality and v is the speed. The net acting force is
(mg − βv)k. By Newton’s second law of motion we have
d2 r
= (mg − βv)k
dt2
dv
m
= (mg − βv)k
dt
dv
m
= (mg − βv)k
dt
Z
Z
m
dv = dt
mg − βv
46
m
CHAPTER 7. RESISTING MEDIUM
47
Integrating both sides gives
−
m
ln(mg − βv) = t + c1
β
At t = 0, v = v0 so that
−
Thus,
m
ln(mg − βv0 ) = c1
β
m
{ln(mg − βv0 ) − ln(mg − βv)}
β
m
mg − βv0
=
ln
β
mg − βv
t=
and if we write α = β/m, this gives
g − αv
= e−αt .
g − αv0
This implies that
g − αv = (g − αv0 )e−αt
Solving for v gives
v=
and thus the velocity is
g
g
+ (v0 − )e−αt
α
α
g −αt i
+ v0 −
e
k
v=
α
α
The acceleration at any time t is
g −αt
e k
a = −α v0 −
α
hg
If z is the distance travelled after time t, then
dz
g g −αt
= + v0 −
e
dt
α
α
Integrating with respect to t gives
z=
g
1
g
t − (v0 − )e−αt + c2
α
α
α
At t = 0, z = 0 so that
c2 =
and therefore
z=
1
g
(v0 − )
α
α
g
1
g
t + (v0 − )(1 − e−αt )
α
α
α
CHAPTER 7. RESISTING MEDIUM
48
Example 7.2. A projectile is launched with the initial speed v0 at an angle α with the
horizontal. Suppose we assume that due to air resistance the projectile has, acting upon it,
a force equal to −βv where β is a positive constant and v is the instantaneous velocity. Find
(a) the velocity and (b) the position vector at any time t.
Solution
(a) By Newton’s second law of motion,
m
d2 r
= −mgk − βv,
dt2
which may as well be written as
dv
β
+ v = −gk.
dt
m
Using the method of integrating factor, we obtain
β
β
e m t v = −f racmβge m t k + c,
where c is a constant of integration. Now, at t = 0, v = v0 = (v0 cos α)j + (v0 sin α)k.
This gives the value of c and we then write
β
mg
mg
−m
t
k+e
(v0 cos α)j + v0 sin α +
k .
v=−
β
β
(b) Since v = dr/dt, the position vector can be obtained by integration and applying the
initial condition, r = 0 at t = 0, and this leads to
β
mg
m
m −βt m
−m
t
1−e
v0 −
r=
t+ e m −
k.
β
β
β
β
7.3
Exercises
1. A man on a parachute falls from rest and acquires a limiting speed of 25 kph. Assuming
that air resistance is proportional to the instantaneous speed, determine how long it
takes to reach the speed of 20 kph.
2. A mass m moves along a straight line under the influence of a constant force of magnitude F . Assuming that there is a resisting force numerically equal to kv 2 where v is
the instantaneous speed and k is a constant, prove that the distance traveled in going
from speed v1 to v2 is
m
F − kv12
ln
2k
F − kv22
CHAPTER 7. RESISTING MEDIUM
49
3. A particle of mass m moves in a straight line acted upon by a constant resisting force
of magnitude F . If it starts with a speed of v0 (a) how long will it take before coming
to rest and (b) what distance will it travel in this time?
4. A locomotive of mass m travels with constant speed v0 along a horizontal track. (a)
How long will it take for the locomotive to come to rest after the ignition is turned off,
if the resistance to the motion is given by α + βv 2 where v is the instantaneous speed
and α and β are constants, (b) what is the distance traveled?
5. A particle moves along the x-axis acted upon only by a resisting force which is proportional to the cube of the instantaneous speed. The initial speed is v0 and after a time
τ the speed is 12 v0 .
(a) Prove that the speed will be 14 v0 in the time 5τ .
(b) Find the total distance traveled by the particle on reaching the speed of 14 v0 .
7.4
Answers
1. Suppose the paracutist is located at a distance z from the ground. Applying Newton’s
second law gives
dv
m = mg − βv.
dt
This may be written by using the technique of separation of variables as
Z
Z
m
dv = dt.
mg − βv
Integrating and applying the initial condition v = 0 at t = 0 gives
mg
m
ln
= t.
β
mg − βv
The paracutist approaches limiting speed when the limiting accelaeration is zero, that
is, mg − βv = 0 or v = 125/18. The time to reach a speed of 20 km/h is calculated to
be 1.12 seconds.
2. From Newton’s second law of motion,
m
d2 x
= F − kv 2 .
dt2
Integrating and applying the initial conditions, at x = 0, v = v1 , and at x = d, say,
v = v2 , where d is the distance travelled during this time, gives the required expression.
CHAPTER 7. RESISTING MEDIUM
50
3. We write, from Newton’s law,
m
F
d2 x
dv
=− ,
=
2
dt
dt
m
since the applied force is in a direction opposite to that of x. Integrating and applying
the condition v = v0 at t = 0 gives
v = v0 −
F
t.
m
The particle will stop when v = 0, that is, t = mv0 /F . The distance travelled at any
time t is obtained by integration and noting that at t = 0, distance x = 0, and this
gives
1 F 2
t.
x = v0 t −
2 m
Thus the distance travelled after t = mv0 /F is
1 mv02
x=
.
2 F
4. Newton’s law of motion enables us to write
m
dv
= −(α + βv 2 ).
dt
Integrating by using method of separation of variables leads to
Z
Z
m
dv = − dt,
α + βv 2
and at t = 0, v = v0 . This will lead to
! √
!
r
r
β
β
αβ
v = tan arctan
v0 −
t .
α
α
m
By using the rule
tan(A + B) =
tan A + tan B
,
1 − tan A tan B
this may as well be written as
√
!
r " r
α
β
αβ
v=
v0 − tan
t
÷
β
α
m
When v = 0 we write
r
β
v0 − tan
α
r
1+
β
v0 tan
α
√
αβ
t = 0,
m
!#
αβ
t
.
m
√
CHAPTER 7. RESISTING MEDIUM
51
and then get the time taken for the locomotive to stop, as
!
r
m
β
t= √
arctan
v0 .
α
αβ
The distance is obtained by integrating the expression
Z
Z
mv
dv = − dx,
α + βv 2
with the conditon v = v0 when x = 0. The distance travelled when v = 0 is thus
α + βv02
m
ln
.
x=
2β
α
5. (a) Solve the equation
dv
= −kv 3 ,
dt
with the conditions v = v0 at t = 0 as follows:
Z
Z
1
k
dv = −
dt
v3
m
1
k
− 2 =− t+c
2v
m
m
The initial condition will give us c = −1/2v02 , and thus
−
or
k
1
1
=
−
t
−
2v 2
m
2v02
k
1
t=
m
2
1
1
−
v 2 v02
This may be simplified to
v2 =
mv02
.
m + 2ktv02
Given that at t = τ , v = v0 /2 so that we obtain k = 3m/2v02 and required is to
find the value of v when t = 5τ . If we substitute for t = 5τ and k = 3m/2v02 in
the expression for v, we obtain v = v0 /4.
(b) Solve the equation
mv
dv
= −kv 3 ,
dx
CHAPTER 7. RESISTING MEDIUM
52
with the conditions v = v0 at t = 0 as follows:
Z
Z
1
k
dv = −
dx
v2
m
1
k
− =− x+c
v
m
The initial condition will give us c = −1/v0 , and thus
k
1
1
− =− v−
v
m
v0
or
m
x=
k
1
1
−
v v0
.
Now, when v = v0 /4 then
x=
3m 2v02
3m
=
= 2v0
kv0
v0 3m
Chapter 8
Constrained motion
8.1
Introduction
We now consider the motion of a single particle constrained to move on curves or surfaces,
for example, the inclined plane or the inner surfaces of a hemispherical bowl. Just as the
particle exerts a force on the constraint, there will, by Newton’s law, be a reaction force of
the constraint on the particle. This reaction force is often described by giving its component
N and F, normal and parallel to the direction of motion respectively. In most cases which
arise in practice, F is the force due to friction and is taken in a direction opposing the motion.
The normal reaction does no work during the motion of a particle. In these circumstances
if the external forces are conservative, the principle of conservation of energy may be applied.
Problems involving constrained motion can be solved using Newton’s second law of motion to arrive at differential equations for the motion then solving these equations subject
to initial conditions.
8.2
Friction
In the constrained motion of a particle, one of the most important forces resisting motion is
that due to friction. Let R be the magnitude of the normal component of the reaction of the
constraint on the particle of mass m as shown in Figure 8.1. Then it is found experimentally
that the magnitude of the force F due to friction is given by
F = µR
where µ is called coefficient of friction.
Example 8.1. A particle P of mass m, slides without rolling down a frictionless inclined
plane AB of angle α with the horizontal. If it starts from rest at the top A of the incline,
find
(a) the acceleration,
53
CHAPTER 8. CONSTRAINED MOTION
54
Figure 8.1: Particle on an inclined plane with friction.
(b) the velocity, and
(c) the distance traveled after time t.
Solution
(a) Since there is no friction the only forces acting on the particle P are the weight w =
−mgk, and the reaction force of the incline which is given by the normal N . Let e1 and
e2 be unit vectors parallel and perpendicular to the incline, respectively. If we denote
by s the magnitude of the displacement from the top A of the inclined plane, we have
by Newton’s second law
d2
m 2 (se1 ) = (mg sin α)e1 ,
dt
for motion of the particle along the inclined plane, and this gives
d2 s
= g sin α
dt2
Thus the acceleration down the incline at any time t is a constant equal to g sin α
(b) Since the speed v is given by v = ds/dt, (8.1) may be written as
dv
= g sin α
dt
On integration we have
v = (g sin α)t + c1
(8.1)
CHAPTER 8. CONSTRAINED MOTION
55
Using the initial conditions at t = 0, v = 0 we obtain c1 = 0, so that the speed at any
time t is
v = (g sin α)t
(8.2)
The velocity ve1 = (g sin α)te1 which has magnitude (g sin α)t in the direction e1 down
the incline.
(c) Since v = ds/dt, (8.2) can be written as
ds
= (g sin α)t + c1
dt
Integrating again with respect to t we obtain
1
s = (g sin α)t2 + c2
2
Using the initial condition s = 0 at t = 0 we have c2 = 0, so that the distance traveled is
1
s = (g sin α)t2
2
(8.3)
Example 8.2. A particle P of mass m, slides without rolling down a frictionless inclined
plane AB of length l and angle α with the horizontal. If it starts from rest at the top A of
the incline, find
(a) the time t taken for the particle to reach the bottom B of the incline, and
(b) the speed of the particle at B.
Solution
(a) Since s = l at B, the time t to reach at the bottom from (8.3) is given by
1
l = (g sin α)t2
2
or
t2 =
2l
g sin α
This implies that
s
t=
2l
g sin α
(b) The speed at B is given from (8.3) by
v = (g sin α)t,
CHAPTER 8. CONSTRAINED MOTION
56
where t is the time the particle takes to reach the bottom B. In this case the time taken
is t obtained in (a) above. So we have
v = (g sin α)t
s
= (g sin α)
=
2l
g sin α
p
2lg sin α
Example 8.3. A particle P of mass m, slides without rolling down an inclined plane AB
of angle α with the horizontal and coefficient of friction µ. If it starts from rest at the top
A of the incline, find
(a) the acceleration,
(b) the velocity, and
(c) the distance traveled after time t.
Solution
(a) In this case there is in addition to the weight and normal reaction acting on the particle
P , a frictional force F directed up the incline and with magnitude µN = µmg cos α By
Newton’s law of motion,
m
d2
(se1 ) = (mg sin α)e1 − (µmg cos α)e1
dt2
(8.4)
This may be written as
d2 s
= g(sin α − µ cos α)
dt2
Thus the acceleration down the incline has the constant magnitude g(sin α − µ cos α)
provided g(sin α > µ cos α) or tan α > µ, otherwise the friction will be so great that the
particle will not move at all.
(b) Replacing d2 s/dt2 by dv/dt in (8.4) and integrating with respect to t, we find that the
speed is given as
v = g(sin α − µ cos α)t
(8.5)
(c) Replacing v with ds/dt in (8.5) and integrating with respect to t we find that the distance
s is given by
1
s = g(sin α − µ cos α)t2
2
CHAPTER 8. CONSTRAINED MOTION
57
Example 8.4. An object slides on a surface of ice along the horizontal straight line OA. At
a certain point in its path the speed is v0 and the object then comes to rest after traveling
a distance x0 . Prove that the coefficient of friction is
µ=
v02
.
2gx0
Solution
Let x be the instantaneous distance of the object of mass m from O and suppose that at
t = 0, x = 0, dx/dt = v0 . There are three forces acting on the object, namely the weight
W = mg, the normal reaction N of the surface on the object and the frictional force F . By
Newton’s second law we have, if v is the instantaneous speed.
m
dv
i=W+N+F
dt
But since N = −W and the magnitude of F is F = µN = µmg so that F = −µmgi, we
have
dv
m i = −µmgi
dt
or
dv
= −µg
(8.6)
dt
On integrating (8.6) and using the fact that v = v0 at t = 0, v = v0 − µgt or
dx
= v0 − µgt
dt
(8.7)
Integrating again, using the fact that x = 0 at t = 0, we find
1
x = v0 t − µgt2
2
From (8.7) we see that the object comes to rest when v0 − µgt = 0 or
t=
v0
µg
Substituting this value of t into equation for x
after it has traveled distance x0 we obtain
v0
x0 = v 0
−
µg
v2
= 0
2µg
and therefore
µ=
and noting that the particle comes to rest
1
µg
2
v02
2gx0
v0
µg
2
CHAPTER 8. CONSTRAINED MOTION
8.3
58
Exercises
1. A weight of 50 kg slides from rest down a 60o incline of length 60 m starting from the
top. Neglecting friction, (a) how long will it take to reach the bottom of the incline
and (b) what is the speed with which it reaches the bottom?
2. Work out problem 1 above taking the coefficient of friction to be 0.3.
3. An object is thrown up a smooth incline of angle α and length l starting from the
bottom, so as to just reach the top.
(a) With what speed should the object be thrown?
(b) What is the time taken to reach the top?
4. If it takes a time t for an object starting from speed v0 on an icy surface to come to
rest, prove that the coefficient of friction is v0 /gt.
5. A mass m rests on a horizontal piece of wood. The wood is tilted upward until the
mass m just begins to slide. If the angle which the wood makes with the horizontal at
that instant is α, prove that the coefficient of friction is µ = tan α.
8.4
Answers
1. Time t = 3.76 seconds and speed at the bottom of the incline is 31.9 metres per second.
2. Time t = 4.14 seconds and speed at the bottom of the incline is 29.1 metres per second.
p
√
3. v0 = 2gl sin α and t = 2l/g sin α.
4. By Newton’s second law of motion,
m
d2 s
= mg sin α − µmg cos α,
dt2
where s is the distance along the inclined plane. Now, when the particle is just about
to slide, mg sin α − mgµ cos α = 0 or µ = tan α, as required.
Chapter 9
Simple harmonic motion
9.1
Introduction
A rectilinear simple harmonic motion is one in which a particle moves along a straight line
under a force which is always directed towards a fixed point O in that line, the magnitude
being proportional to the particle’s displacement from O.
A mass m lies on a frictionless horizontal table indicated by the x-axis and is attached
to one end of a spring of negligible mass and unstretched length l whose other end is fixed
at E. If the mass is given a displacement along the x-axis and released, it will vibrate back
and forth about the equilibrium position O. Now when the spring is of length l + x there is
a force tending to restore m to equilibrium position, as shown in Figure 9.1. By Hooke’s law
this restoring force is proportional to the stretch x and is given by
FR = −kxi
where FR stands for restoring force and k is a constant of proportionality which is called
spring constant, elastic constant, stiffness factor or modulus of elasticity.
By Newton’s second law of motion we have
m
d2 x
i = −kxi
dt2
or
d2 x
+ kx = 0
(9.1)
dt2
This vibrating system is called the simple harmonic oscillator or linear harmonic
oscillator. This type of motion is often called simple harmonic motion.
Equation (9.1) can be written as
m
d2 x
k
+ x=0
2
dt
m
or
d2 x
+ ω2x = 0
dt2
59
CHAPTER 9. SIMPLE HARMONIC MOTION
60
Figure 9.1: A horizontal spring-mass system oscillating about O with amplitude A.
where ω 2 = k/m.
9.2
Amplitude, period and frequency
Now, let us solve the differential equation
d2 x
+ ω2x = 0
dt2
(9.2)
with initial conditions, x = A, dx/dt = 0 at t = 0. Since the derivative of an exponential
function is a constant multiple of itself, we try a solution of the form
x = eαt
where α is a constant to be determined. Then
dx
d2 x
= αeαt and 2 = α2 eαt
dt
dt
Substituting these values in (9.2) we obtain
α2 eαt + ω 2 eαt = 0
This implies that
(α2 + ω 2 )eαt = 0
Since eαt ̸= 0, we have α2 + ω 2 = 0 which is an algebraic equation in α. The solution of this
equation is
α = ±iω
CHAPTER 9. SIMPLE HARMONIC MOTION
61
Thus the two linearly independent solutions of the given differential equation are
eiωt and e−iωt
The general solution is thus
x = C1 eiωt + C2 e−iωt = c1 cos ωt + c2 sin ωt
where c1 and c2 are arbitrary constants to be determined from initial conditions. Now, at
t = 0, x = A, the amplitude, so that
A = c1 cos 0 + c2 sin 0.
This implies that c1 = A and therefore
x = A cos ωt + c2 sin ωt
Now,
dx
= −ωA sin ωt + c2 ω cos ωt
dt
At t = 0, dx/dt = 0 so that
= −ωA sin 0 + c2 ω cos 0
This gives c1 = 0 since ω ̸= 0 and therefore
x = A cos ωt.
Since cos ωt varies between −1 and 1, x varies between −A and A. Thus the mass
oscillates between −A and A. The amplitude of the motion is the greatest distance from
the equilibrium position. In this case it is A. The period of the motion is the time for one
complete oscillation, sometimes called a cycle. In this case it is the time taken from A to -A
and then back to A.
Let T denote the period then
r
m
2π
= 2π
T =
ω
k
The frequency of the motion denoted by f is the number of complete oscillations per unit
time i.e.,
r
1
ω
1
k
f= =
=
T
2π
2π m
We may write the solution x = c1 cos ωt + c2 sin ωt in the form
q
c2
2
2 p c1
cos ωt + p 2
sin ωt
x = c1 + c2
c21 + c22
c1 + c22
CHAPTER 9. SIMPLE HARMONIC MOTION
62
Let tan ϕ = c2 /c1 . Then
sin ϕ = p
Thus
c
p 1
and
cos
ϕ
=
c21 + c22
c21 + c22
c2
q
x = c21 + c22 (cos ϕ cos ωt + sin ϕ sin ωt) = C cos(ωt − ϕ)
In this representation for x, the amplitude is C, whereas the period and frequency remain
the same. The angle ϕ is called the phase angle or epoch. It is chosen so that 0 ≤ ϕ ≤ π. If
ϕ = 0 we obtain the previous expression.
9.3
Energy of the harmonic oscillator
A simple harmonic oscillator is an oscillator that is neither driven nor damped. It consists
of a mass m, which experiences a single force, F, which pulls the mass in the direction of the
point x = 0 and depends only on the mass’s position x and a constant k. Balance of forces
(Newton’s second law) for the system is
F = ma = m
d2 x
= mẍ = −kx.
dt2
Let T be the kinetic energy and V the potential energy of a particle moving with simple
harmonic motion. Then the total energy of the simple harmonic oscillator is E = T + V .
We already know that T = 12 mv 2 and V = 12 kx2 , where m is the mass of the particle, v the
speed of the particle, x the distance of the particle from the equilibrium point, and k is the
spring constant. Thus, from the principle of conservation of energy,
1
1
E = mv 2 + kx2 .
2
2
Example 9.1. A particle P of mass 2 units moves along the x-axis attracted toward origin
O by a force whose magnitude is numerically equal to 8x. If it is initially at rest at x = 20,
find
(a) the differential equation and initial conditions describing the motion,
(b) the position of the particle,
(c) the speed and velocity, at any time t,
(d) the amplitude, period and frequency of the vibration.
Solution
CHAPTER 9. SIMPLE HARMONIC MOTION
63
(a) Let r = xi be the position vector of P . The acceleration of P is
d2
d2 x
(xi)
=
i.
dt2
dt2
The net force acting on P is −8xi. Then, by Newton’s second law,
m
d2 x
i = −8xi
dt2
or
d2 x
+ 4x = 0,
(9.3)
dt2
which is the required differential equation describing the motion. The initial conditions
are x = 20, dx/dt = 0 at t = 0.
(b) The general solution of (9.3) is
x = c1 cos 2t + c2 sin 2t,
where c1 and c2 are arbitrary constants. At t = 0, x = 20 so that
20 = c1 cos 0 + c2 sin 0.
This implies that c1 = 20 and therefore
x = 20 cos 2t + c2 sin 2t
Differentiating with respect to t we have
dx
= −40 sin 2t + 2c2 cos 2t
dt
At t = 0, dx/dt = 0, so we have
0 = −40 sin 2t + 2c2 cos 2t
This implies that c2 = 0 and thus
x = 20 cos 2t
which gives the position of the particle at any time t.
(c) From (9.4)
dx
= −40 sin 2t,
dt
which gives the speed of the particle at any time t. The velocity is given by
dx
i = −40 sin 2ti.
dt
(9.4)
CHAPTER 9. SIMPLE HARMONIC MOTION
64
(d) The amplitude is 20, period = 2π/ω = 2π/2 = π, and frequency = π1 .
Example 9.2. Work out the previous example taking P to be initially at 20 but moving (a)
to the right with speed 30, (b) to the left with speed 30. Find the amplitude, period and
frequency in each case.
Solution
(a) In this case, the condition dx/dt = 0 at t = 0 is replaced by dx/dt = 30 at t = 0. Then
one obtains c2 = 15 and therefore
x = 20 cos 2t + 15 sin 2t
which gives the position P at any time. This may be written as
√
20
15
cos 2t + √
sin 2t
x = 202 + 152 √
202 + 152
202 + 152
3
4
cos 2t + sin 2t
= 25
5
5
= 25 cos(2t − ϕ),
where cos ϕ =
4
5
and sin ϕ = 35 .
Since the cosine varies between −1 and +1 the amplitude is 25. The period and frequency
are the same as before.
(b) In this case the condition dx/dt = 0 at t = 0 is replaced by dx/dt = −30 at t = 0. Then,
one obtains c2 = −15 and therefore
x = 20 cos 2t − 15 sin 2t,
which gives the position P at any time. This may be written as
√
20
15
2
2
x = 20 + 15 √
cos 2t − √
sin 2t
202 + 152
202 + 152
3
4
= 25
cos 2t + sin 2t
5
5
= 25 cos(2t − ψ),
where cos ψ = 45 and sin ψ = − 53 . The amplitude, period and frequency are the same as
in (a). The only difference is the phase angle. The relationship between ψ is ψ = ϕ + π.
This is often described by saying that the two motions are 180o out of phase with each
other.
Example 9.3. An object of mass 20 kg moves with simple harmonic motion on the x-axis.
Initially it is located at a distance 4 metres away from the origin x = 0 and has velocity 15
m/s and acceleration 100m/s2 directed toward x = 0. Find
CHAPTER 9. SIMPLE HARMONIC MOTION
65
(a) the position at any time ,
(b) the amplitude, period and frequency of oscillation, and
(c) the force on the object when t =
π
10
seconds.
Solution
(a) If x denotes the position of the object at time t then the initial conditions are x = 4,
dx/dt = −15, d2 x/dt2 = −100 at t = 0. Now, for simple harmonic motion, we have
x = c1 cos ωt + c2 sin ωt
At t = 0, x = 4, giving c1 = 4. Thus
x = 4 cos ωt + c2 sin ωt
Differentiating this gives
dx
= −4ω sin ωt + ωc2 cos ωt
dt
At t = 0, dx/dt = −15, giving ωc2 = −15. Thus
dx
= −4ω sin ωt − 15 cos ωt
dt
Differentiating this again gives
d2 x
= −4ω 2 cos ωt + 15ω sin ωt
dt2
At t = 0, d2 x/dt2 = −100 giving ω 2 = 25. This then gives ω = 5 and c2 = −3. Thus
x = 4 cos 5t − 3 sin 5t
(9.5)
(b) Equation (9.5) may be rewritten as
√
20
15
2
2
x = 20 + 15 √
cos 5t − √
sin 5t
202 + 152
202 + 152
4
3
=5
cos 5t + sin 5t
5
5
= 5 cos(5t − ϕ),
where cos ϕ = 45 and sin ϕ = − 53 . The amplitude is 5 m, period is 2π/5 seconds and
5
frequency is 2π
Hz or cycles per second.
CHAPTER 9. SIMPLE HARMONIC MOTION
66
(c) We know that F = ma. From (9.5), we obtain the acceleration by differentiating twicw
with respect to t. The answer can easily be obtained as 1500 N.
Example 9.4. Derive the equation
1
1
E = mv 2 + kx2 ,
2
2
for the simple harmonic oscillator.
Solution
The differential equation for the simple harmonic oscillator is obtained by Newton’s second
law of motion, as
d2 x
m 2 = −kx
dt
dv
= −kx
mv
dx
mvdv = −kxdx.
Integrating gives
1
1 2
mv = − kx2 + E,
2
2
which may be written as
1
1
E = mv 2 + kx2 .
2
2
9.4
Exercises
1. A particle of mass 12 grams moves along the x-axis attracted toward the point O on it
by a force in dynes which is numerically equal to 60 times its instantaneous distance
x (in cm) from O. If the particle starts from rest at x = 10 cm, find the (a) position
at any time t, (b) period and (c) frequency of motion.
2. (a) If the particle of problem 1 above starts at x = 10 cm with a speed toward O of 20
cm/s, determine its amplitude, period and frequency. (b) Determine when the particle
reaches O for the first time.
3. A particle moves with simple harmonic motion in a straight line. Its maximum speed
is 15 m/s and its maximum acceleration is 135 m/s2 . Find the period and frequency
of the motion.
4. A particle moves with simple harmonic motion. If its acceleration at distance
l from
p
the equilibrium position is a, prove that the period of the motion is 2π l/a.
5. A particle moving with simple harmonic motion has speed 3 cm/s and 4 cm/s at
distances 8 cm and 6 cm, respectively, from the equilibrium position. Find the period
of the motion.
CHAPTER 9. SIMPLE HARMONIC MOTION
67
6. When a mass m1 hanging from the lower end of a vertical spring is set into motion, it
oscillates with period T . Prove that the period T1 when a mass m2 is added is given
as
r
m2
T1 = T 1 +
m1
9.5
Answers
1. Let r = xi be the position vector of the particle, say, P . The acceleration of P is
d2
d2 x
(xi)
=
i.
dt2
dt2
The net force acting on P is −60xi. Then by Newton’s second law,
12
d2 x
i = −60xi
dt2
or
d2 x
+ 5x = 0,
dt2
which is the required differential equation describing the motion. The initial conditions
are x = 10, dx/dt = 0 at t = 0.
(a) The general solution of this equation is
√
√
x = c1 cos 5t + c2 sin 5t
where c1 and c2 are arbitrary constants. Since at t = 0, x = 20, then
10 = c1 cos 0 + c2 sin 0.
This implies that c1 = 10 and therefore
√
√
x = 10 cos 5t + c2 sin 5t.
Differentiating with respect to t we have
√
√
√
√
dx
= −10 5 sin 5t + 5c2 cos 5t
dt
At t = 0, dx/dt = 0, so we have
0 = −40 sin 2t + 2c2 cos 2t.
This implies that c2 = 0 and thus
x = 10 cos
√
5t
which gives the position of the particle at any time t, with amplitude 10 cm.
CHAPTER 9. SIMPLE HARMONIC MOTION
68
√
(b) The period is given as T = 2π/ω = 2π/ 5 seconds.
√
(c) The frequency =
5
2π
hertz.
2. (a) The differential equation describing this motion is
d2 x
+ 5x = 0,
dt2
with initial conditions x = 10, dx/dt = −20 at t = 0. The general solution of this
equation is
√
√
x = c1 cos 5t + c2 sin 5t
where c1 and c2 are arbitrary constants. Since at t = 0, x = 20, then
10 = c1 cos 0 + c2 sin 0
This implies that c1 = 10 and therefore
√
√
x = 10 cos 5t + c2 sin 5t
Differentiating with respect to t we have
√
√
√
√
dx
= −10 5 sin 5t + 5c2 cos 5t
dt
At t = 0, dx/dt = −20, so we have
√
√
√
√
−20 = −10 5 sin 5t + 5c2 cos 5t
√
This implies that c2 = −4 5 and thus
√
√
√
x = 10 cos 5t − 4 5 sin 5t
which gives the position of the particle at any time t. Now, this equation may be
rewritten as


√
q
√
√
√
10
4 5
x = 102 + (4 5)2  q
cos 5t − q
sin 5t
√
√
102 + (4 5)2
102 + (4 5)2
!
√
√
√
√
10
4 5
√ cos 5t − √ sin 5t
= 180
6 5
6 5
!
√
√
√
√
5
2
cos 5t − sin 5t
=6 5
3
3
√
√
= 6 5 cos( 5t + ϕ)
√
√
√
where cos ϕ = 35 and sin ϕ = 32 . The amplitude is 6 5 and the period is 2π/ 5.
CHAPTER 9. SIMPLE HARMONIC MOTION
69
√
√
(b) The particle reaches O for the first time, when x = 0, that is, 6 5 cos( 5t + ϕ) =
0. Solving for t gives t = 0.34 seconds.
3. Since the particle moves with simple harmonic motion, its motion is given by x =
A cos ωt. The speed at any time t is given as dx/dt = −Aω sin ωt and this is maximum
when sin ωt = 1 or ωt = π/2. Now, if maximum speed is 15 m/s, then we write
−Aω = −15 or A = 15/ω so that
x=
15
cos ωt.
ω
The speed is then dx/dt = −15ω sin ωt and the acceleration is d2 x/dt2 = −15ω 2 cos ωt.
This acceleration is maximum when cos ωt = 1. We then write −15ω 2 = −135 or
ω 2 = 9, and therefore,
15
cos 3t = 5 cos 3t.
x=
3
This gives the position of the particle at any time t, with amplitude 5, and the period
is given as T = 2π/ω = 2π/3 seconds. Frequency is f = ω/2π = 3/2π.
4. Since the particle moves with simple harmonic motion, its motion is given by x =
A sin ωt. Now, for a particle of mass m attached to a spring of constant k, by Newton’s
law of motion,
d2 x
d2 x
k
m 2 = kx or
= − x.
2
dt
dt
m
p
2
This means that ω = k/m or ω = k/m. Now, when the acceleration is a and the
distance is l, then we write
ma = −kl
Thus, ω =
or
k
a
=− .
m
l
p
p
k/m = a/l, and the period is given as
r
2π
l
T =
= 2π
.
ω
a
5. Since the particle moves with simple harmonic motion, its motion is given by x =
A sin ωt. Now, for a particle of mass m attached to a spring of constant k, ω 2 = k/m.
By Newton’s second law of motion,
d2 x
dt2
d2 x
dt2
dv
v
dx
vdv
m
= −kx
= −ω 2 x
= −ω 2 x
= −ω 2 xdx
CHAPTER 9. SIMPLE HARMONIC MOTION
70
Integrating gives
v2
1
= − ω 2 x2 + c,
2
2
where c is a constant. Applying the initial conditions v = 3 at x = 8 and v = 4 at
x = 6 gives ω = 21 . The period of motion is then T = 2π/ω = 2π/0.5 = 4π seconds.
6. Since the particle moves with simple harmonic motion, its motion is given by x =
A sin ωt. Now, for a particle of mass m1 attached to a spring of constant k, by Newton’s
second law of motion,
d2 x
d2 x
k
m1 2 = kx or 2 = − x.
dt
dt
m
p
This means that ω 2 = k/m, and therefore the period is T = 2π/ω = 2π m1 /k. When
the mass m2 is now added, then
(m1 + m2 )
d2 x
d2 x
k
x.
=
kx
or
=−
2
2
dt
dt
m1 + m2
2
This
p means that ω = k/(m1 + m2 ), and therefore the period is T = 2π/ω =
2π (m1 + m2 )/k. Eliminating k in these equations gives the desired result.
Chapter 10
The 2D- and 3D-harmonic oscillators
10.1
Introduction
Suppose a particle of mass m moves in the xy-plane under the influence of a force field F
given by
F = −k1 xi − k2 yj
where k1 and k2 are constants. Here the position vector r is given by
r = xi + yj.
By Newton’s second law of motion we have
d2 r
=F
dt2
d2 (xi + yj)
= −k1 xi − k2 yj,
m
dt2
m
from which we obtain the equations of motion of the form
d2 x
d2 y
m 2 = −k1 x and m 2 = −k2 y.
dt
dt
We may then write this as
d2 x
d2 y
2
+
w
x
=
0
and
+ w22 x = 0,
1
dt2
dt2
where ω12 = k1 /m and ω22 = k2 /m. Solutions of these equations are, respectively,
x = c1 cos ω1 t + c2 sin ω1 t
and
y = c3 cos ω2 t + c4 sin ω2 t
71
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
72
where c1 , c2 , c3 , c4 are constants to be determined from the initial conditions. The mass m
subjected to the force field above is often called a two-dimensional harmonic oscillator. These
ideas can easily be extended to three dimensional harmonic oscillator of mass m which is
subject to a force field given be
F = −k1 xi − k2 yj − k3 zk,
where k1 , k2 , and k3 are positive constants.
10.2
Worked examples
Example 10.1. A particle moves in the xy-plane in a force field given by F = −kxi − kyj.
Prove that in general it will move in an elliptical path.
Solution
If the particle has mass m, then by Newton’s second law of motion we have
m
or
d2 r
=F
dt2
d2 (xi + yj)
= −kxi − kyj.
dt2
Since the position vector r in this case is r = xi + yj, then the equations of motion are
m
m
d2 x
d2 y
=
−kx
and
m
= −ky.
dt2
dt2
If ω 2 = k/m then equations of motion become
d2 x
+ ω 2 x = 0 and
dt2
d2 y
+ ω 2 y = 0.
dt2
The solutions are
x = A1 cos ωt + B1 sin ωt and y = A2 cos ωt + B2 sin ωt.
Assume the initial conditions r = ai + bj and dr/dt = v1 i + v2 j at t = 0. Now, using t = 0
and r = ai + bj gives A1 = a and A2 = b, and thus
x = a cos ωt + B1 sin ωt and y = b cos ωt + B2 sin ωt
Also
dx
= −aω sin ωt + ωB1 cos ωt and
dt
dy
= −bω sin ωt + ωB2 cos ωt
dt
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
73
Using the intial condition dx
= v1 and dy
= v2 at t = 0, we have ωB1 = v1 and ωB2 = v2 , or
dt
dt
B1 = v1 /ω and B2 = v2 /ω. If we let c = v1 /ω and d = v2 /ω, then the solutions become
x = a cos ωt + c sin ωt
(10.1)
y = b cos ωt + d sin ωt.
(10.2)
and
We now solve for sin ωt and cos ωt by multiplying (10.1) by d and (10.2) by c and
subtracting, to get (ad − bc) cos ωt = dx − cy or
cos ωt =
dx − cy
ad − bc
Similarly by multiplying (10.1) by b and(10.2) by a and subtracting we obtain (ad −
bc) sin ωt = ay − bx or
ay − bx
sin ωt =
ad − bc
By Pythagoras’ theorem, we obtain
(dx − bc)2 (ay − bx)2
+
= cos2 ωt + sin2 ωt = 1,
(ad − bc)2
(ad − bc)2
and thus we obtain
(d2 + b2 )x2 − 2(ab + cd)xy + (a2 + c2 )y 2 = (ad − bc) = 1.
Now, from Algebra, the quadratic equation Ax2 + Bxy + Cy 2 = D, where A > 0, C > 0,
and D > 0 is elliptic if B 2 − 4AC < 0, parabolic if B 2 − 4AC = 0, and hyperbolic if
B 2 − 4AC > 0. In this case we have A = d2 + b2 , B = −2(ab + cd), and C = a2 + c2 ,
and therefore B 2 − 4AC = 4(ab + cd)2 − 4(d2 + b2 )(a2 + c2 ) = 4(−a2 d2 + 2abcd − b2 c2 ) =
−4(a2 d2 − 2abcd + b2 c2 ) = −4(ad − bc)2 < 0. Thus this particle will in general move in an
elliptical path.
Example 10.2. Find the potential energy for the three dimensional harmonic oscillator.
Solution
In this we have
F = −k1 xi − k2 yj − k3 zk,
where k1 , k2 , and k3 are positive constants. The force field F is conservative if and only if
∇ × F = 0. In this case therefore we write
∇×F=
i
j
k
∂
∂x
∂
∂y
∂
∂z
−k1 x −k2 y −k3 z
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
74
This gives ∇ × F = 0 and so the force field is conservative, and therefore there exists a
potential V such that F = −∇V or
−k1 xi − k2 yj − k3 zk = −
or
∂V
= k1 x,
∂x
∂V
∂v
∂V
i−
j−
k
∂x
∂y
∂z
∂V
= k2 y,
∂y
∂V
= k3 z
∂z
(10.3)
Integrating the first equation of (10.3) with respect to x, keeping y and z constant we
obtain
1
(10.4)
V = k1 x2 + g(y, z)
2
Differentiating (10.4) with respect to y and equating to the second equation in (10.3) we
have
∂V
∂g
=
= k2 y
(10.5)
∂y
∂y
so that
∂g
∂y
= k2 y. Now, integrating (10.5) with respect to y keeping z constant we have
1
g(y, z) = k2 y 2 + f (z)
2
So (10.4) becomes
1
1
(10.6)
V = k1 x2 + k2 y 2 + f (z)
2
2
Now differentiating (10.6) with respect to z and equating to the last equation in (10.3)
we have
∂V
df
=
= k3 z
∂z
dz
Integrating gives f = 12 k3 z 2 + c, and thus the potential is
1
1
1
V = k1 x2 + k2 y 2 + k3 z 2 + c,
2
2
2
where c is taken to be zero.
10.3
Exercises
1. A particle of mass 2 units moves in the xy-plane attracted to the origin with a force
given by F = −18xi − 50yj. At t = 0 the particle is placed at the point (3, 4) and given
a velocity of magnitude 10 in a direction perpendicular to the x-axis.
(a) Find the position and velocity of the particle at any time t.
(b) Find the total energy of the particle.
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
75
2. A two dimensional harmonic oscillator of mass 2 units has potential energy given by
V = 8(x2 + 4y 2 ). If the position vector and velocity of the oscillator at the time t = 0
are given respectively by r0 = 2i − j and v0 = 4i + 8j
(a) Find its position and velocity at any time t > 0, and
(b) determine the period of motion.
3. A particle of mass m moves in 3-dimensional force field whose potential is given by
V = 21 k(x2 + y 2 + z 2 )
(a) Prove that if the particle is placed at arbitrary point in space other than the
origin, then it will return to the point after some period of time. Determine this
period.
(b) Is the velocity on returning to the starting point the same as the initial velocity?
Explain.
4. A particle moves in the xy-plane so that its position is given by x = A cos ωt,y =
B cos 2ωt. Prove that it describes an arc of a parabola.
5. The position of a particle moving in the xy-plane is described by the equations
d2 x
= −4y
dt2
and
d2 y
= −4x
dt2
At t = 0, the particle is at rest at the point (6, 3). Find (a) its position, and (b) its
velocity, at any time t.
10.4
Answers
1. (a) The external force acting on the particle is F = −18xi − 50yj. By Newton’s
second law we have
d2 r
2 2 = −18xi − 50yj,
dt
where r = xi + yj. We then obtain the equations
d2 x
+ 9x = 0 and
dt2
d2 y
+ 25y = 0
dt2
The solutions to these equations are x = A1 cos 3t + B1 sin 3t and y = A2 cos 5t +
B2 sin 5t. Now, at t = 0, x = 3 and y = 4, so that A1 = 3 and A2 = 4. Thus,
x = 3 cos 3t + B1 sin 3t and y = 4 cos 5t + B2 sin 5t.
Also
dx
= −9 sin 3t + B1 cos 3t and
dt
dy
= −20 sin 5t + B2 cos 5t
dt
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
76
At t = 0, dx
= 0 and dy
= 10 and so B1 = 0 and B2 = 2. Thus x = 3 cos 3t and
dt
dt
y = 4 cos 5t + 2 sin 5t. Hence the position at any time t is
r = 3 cos 3ti + (4 cos 5t + 2 sin 5t)j.
The velocity v is given by
v=
dr
= −9 sin 3ti + (10 cos 5t − 20 sin 5t)j
dt
(b) The kinetic energy at t = 0, v = 10 is
1
1
K.E. = mv 2 = × 2 × 102 = 100
2
2
Since F = −18xi − 50yj is a conservative force field, there is a scalar potential V
such that F = −∇V , that is,
∂V
= 18x and
∂x
∂V
= 50y
∂y
This means that ∂V /∂x = 18x so that V = 9x2 + g(y). Also,
∂V
∂g
=
= 50y,
∂y
∂y
and thus g = 25y 2 so that
V = 9x2 + 25y 2 ,
where the constant of integration is zero. Now, by the principle of conservation of
energy we have that the sum of potential and kinetic energy equals E, a constant,
that is, E = 9(3)2 + 25(4)2 + 100 = 581.
2. (a) The force acting on the particle in this force field is F = −∇V . By Newton’s
second law we have
d2 r
2 2 = −16xi − 64yj,
dt
where r = xi + yj. Thus
d2
2 2 (xi + yj) = −16xi − 64yj
dt
Hence we obtain the equations
d2 x
+ 8x = 0 and
dt2
d2 y
+ 32y = 0
dt2
The solutions to these equations are, respectively,
√
√
√
√
x = A1 cos 2 2t + B1 sin 2 2t and y = A2 cos 4 2t + B2 sin 4 2t
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
77
At t = 0, r0 = 2i − j,√that is, x0 = 2√and y0 = −1. Thus√A1 = 2 and A√2 = −1 and
therefore x = 2 cos 2 2t + B1 sin 2 2t and y = − cos 4 2t + B2 sin 4 2t. Also
√
√
√
√
dx
= −4 2 sin 2 2t+2 2B1 cos 2 2t and
dt
√
√
√
√
dy
= −4 2 sin 2 2t+4 2B2 cos 4 2t
dt
√
At t=0
v
=
4i
+
8j,
that
is,
dx/dt
=
4
and
dy/dt
=
8.
Thus
4
=
2
2B1 or
0
√
√
√
B1 = 2 and 8 = 4 2B2 or B2 = 2, and therefore
√
√
√
√
√
√
x = 2 cos 2 2t + 2 sin 2 2t and y = − cos 4 2t + 2 sin 4 2t.
Thus the position r at any time is
√
√
√
√
√
r = (2 cos 2 2t + 2 sin 2 2t)i + (− cos 4 2t + 2 sin 4 2t)j
and the velocity v at any time is
v=
√
√
√
√
√
√
dr
= (4 cos 2 2t − 4 2 sin 2 2t)i + (4 2 sin 4 2t + 8 cos 4 2t)j
dt
(b) Given that
√
√
√
√
√
r = (2 cos 2 2t + 2 sin 2 2t)i + (− cos 4 2t + 2 sin 4 2t)j
This may be written as
√
√
√
√
r = 6 cos(2 2t − ψ1 )i + 3 cos(4 2t − ψ2 )j,
q
1
2
√
√
where sin ψ1 = 3 , and cos ψ1 = 6 , sin ψ2 = − 23 ,and cos ψ2 =
is T =
2π
√
2 2
√
=π
√1 .
3
The period
2
.
2
3. The force acting is F = −∇V , where
F = −kxi − kyj − kzk
By Newton’s second law we have
m
d2 r
= −kxi − kyj − kzk,
dt2
where r = xi + yj + zk. Thus we obtain the equations
d2 x
= −ω 2 x,
dt2
where ω 2 =
k
.
m
d2 y
= −ω 2 y,
dt2
d2 z
= −ω 2 z
dt2
The general solutions of these equations are
x = A1 cos ωt + B1 sin ωt y = A2 cos ωt + B2 sin ωt z = A3 cos ωt + B3 sin ωt
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
78
or r = (A1 cos ωt + B1 sin ωt)i + (A2 cos ωt + B2 sin ωt)j + (A3 cos ωt + B3 sin ωt)k. This
may be written as
(A1 i + A2 j + A3 k) cos ωt + (B1 i + B2 j + B3 k) sin ωt = A cos ωt + B sin ωt,
where A = A1 i+A2 j+A3 k and B = B1 i+B2 j+B3 k. At t = 0, r0 = A cos 0+B sin 0 =
A. The position will again be r0 = A when ωt = 2π or t = 2π/ω. At any time t we
have
r = A cos ωt + B sin ωt
and
dr
= −ωA sin ωt + ωB cos ωt,
dt
and when t = 2π/ω the velocity is
dr
2π
2π
= −ωA sin ω
+ ωB cos ω
= ωB
dt
ω
ω
The initial velocity is v(0) = −ωA sin 0 + ωB cos 0 = ωB, and therefore the velocity
on returning to the starting point is the same as the initial velocity.
4. Given that x = a cos ωt and y = b cos 2ωt and it should be noted that cos 2ωt =
2 cos2 ωt − 1. Then, we eliminate cos ωt from both equations and obtain
!
2
x
−1
y=b 2
a
This may be written as
x2 =
a2
(y + b)
b
2
and this is an equation of a parabola with vertex at (0, −b) and focus at (0, a8b ).
5. Given that
d2 x
= −4y
dt2
and
d2 y
= −4x
dt2
From the first equation, differentiating twice gives
d4 x
d2 y
=
−4
,
dt4
dt2
and using the second equation gives
d4 x
= −4(−4x) = 16x
dt4
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
79
The solution to this equation is
x = A cos 2t + B sin 2t + Ce2t + De−2t
At t = 0, x = 6 so that
A+C +D =6
(10.7)
Also,
dx
= −2A sin 2t + 2B cos 2t + 2Ce2t − 2De−2t
dt
and at t = 0, dx/dt = 0 so that
B+C −D =0
(10.8)
Again,
d2 x
= −4A cos 2t − 4B sin 2t + 4Ce2t + 4De−2t
dt2
and at t = 0,
d2 x
dt2
= −4y = −12 so that
A−C −D =3
(10.9)
Differentiating the third time gives
d3 x
= 8A cos 2t − 8B sin 2t + 8Ce2t + 8De−2t
3
dt
and at t = 0,
d3 x
dt3
= −4 dy
= 0 so that
dt
B−C +D =0
(10.10)
Solving (10.7), (10.8), (10.9), and (10.10) simultaneously we obtain A = 29 , B = 0, and
C = D = 34 . Therefore
x=
9
3
9
3
cos 2t + (e2t + e−2t ) = cos 2t + cosh 2t
2
4
2
4
Now, from the given equations,
y=−
where
dx
= −9 sin 2t + 3 sin 2t and
dt
so that
y=
1 d2 x
,
4 dt2
d2 x
= −18 cos 2t + 6 sinh 2t
dt2
9
3
cos 2t − cosh 2t
2
2
CHAPTER 10. THE 2D- AND 3D-HARMONIC OSCILLATORS
Thus the position vector is given as
3
9
3
9
cos 2t + cosh 2t i +
cos 2t − cosh 2t j
r=
2
2
2
2
Thus the velocity is obtained by differentiating r as
v = (−9 sin 2t + 3 sinh 2t)i − (9 sin 2t + 3 sinh 2t)j
80
Chapter 11
The damped harmonic oscillator
11.1
Introduction
In real oscillators, friction, or damping, slows the motion of the system. Due to frictional
force, the velocity decreases in proportion to the acting frictional force. While simple harmonic motion oscillates with only the restoring force acting on the system, damped harmonic
motion experiences friction. In many vibrating systems the frictional force Ff can be modeled as being proportional to the velocity v of the object: Ff = −cv, where c is called
the viscous damping coefficient. Balance of forces, according to Newton’s second law, for
damped harmonic oscillators is then
d2 x
dx
=m 2.
F = −kx − c
dt
dt
When no external forces are present, i.e., when F = 0, this can be rewritten into the
form
dx
d2 x
+
2γ
+ ω 2 x = 0,
(11.1)
dt2
dt
q
β
k
where ω = m
is called the ’undamped angular frequency of the oscillator’ and γ = 2m
is
called the ’damping ratio’. The value of the damping ratio critically determines the behavior
of the system.
To solve (11.1), we try a solution of the form x = ert , meaning that ẋ = rert and ẍ = r2 ert .
Substituting for these in the given equation gives
r2 ert + 2γrert + ω 2 ert = 0,
or (r2 + 2γr + ω 2 )ert = 0. Now, since ert ̸= 0, we obtain the algebraic equation
r2 + 2γr + ω 2 = 0.
This is a quadratic equation in r and therefore has two roots, given as
p
r1 , r2 = −γ ± γ 2 − ω 2
81
CHAPTER 11. THE DAMPED HARMONIC OSCILLATOR
82
According to this solution, a damped harmonic oscillator can be, as illustrated in Figure
11.1,
1. Overdamped (γ 2 > ω 2 ): The system returns (exponentially decays) to steady state
without oscillating. Larger values of the damping ratio return to equilibrium slower.
2. Critically damped (γ 2 = ω 2 ): The system returns to steady state as quickly as possible without oscillating (although overshoot can occur). This is often desired for the
damping of systems such as doors.
3. Underdamped (γ 2 < ω 2 ): The system oscillates (with a slightly different frequency
than the undamped case) with the amplitude gradually decreasing to zero.
Figure 11.1: Types of damping.
11.2
Cases of damping
In case of over-damped motion, where γ 2 > ω 2 or β 2 > 4mk, there are two real roots to the
quadratic equation and the general solution of the differential equation is
x = e−γt c1 eαt + c2 e−αt ,
p
where α = γ 2 − ω 2 and c1 and c2 are arbitrary constants which can be found by using
initial conditions. The form of this solution is a periodic or dead beat, both exponentially
decaying with time.
In case of critically damped motion, where γ 2 = ω 2 or β 2 = 4mk, there are two identical
real roots. The general solution is of the form
x = e−γt (c1 + c2 t) .
CHAPTER 11. THE DAMPED HARMONIC OSCILLATOR
83
Again the motion is periodic and the system is said to be critically damped.
In case of under-damped or damped oscillatory motion, γ 2 < ω 2 or β 2 < 4mk, the roots
of the quadratic equation are complex, and the general solution is of the form
x = e−γt (c1 sin λt + c2 cos λt) ,
or
x = ce−γt cos(λt − ϕ),
p
p
where λ = γ 2 − ω 2 and c = c21 + c22 is called the amplitude and ϕ is called the phase
angle or epoch. In the first two cases, the damping is so large that no oscillation takes place.
The mass simply returns to the equilibrium position.
In the third case, oscillations occur although the amplitude of these oscillations tends to
decrease with time. The difference in time between two successive maxima (or minima) in
the under-damped motion is called the period of the motion and is given by
T =
2π
4πm
2π
=p
=p
.
λ
gamma2 − ω 2
β 2 − 4km
The period and frequency corresponding to β = 0 are called natural period and natural
frequency, respectively. The period T given by
T =
2π
2π
4πm
=p
=p
λ
γ 2 − ω2
β 2 − 4km
is also equal to two successive values of t for which cos(λt − ϕ) = 1 or cos(λt − ϕ) =
−1. Suppose that the values of x corresponding to tn and tn+1 = tn + T are xn and xn+1
respectively. Then
e−γtn
xn
= −γtn +T = eγT .
xn+1
e
The quantity
ln
xn
xn+1
= γT,
which is a constant, is called the logarithmic decrement.
Example 11.1. Prove that the force F = −kxi acting on a simple harmonic oscillator is
conservative. Hence find the potential energy of the simple harmonic oscillator.
Solution
F is a conservative force if and only if ∇ × F = curl F = 0. We write
∇×F=
i
j
k
∂
∂x
∂
∂y
∂
∂z
−kx
0
0
CHAPTER 11. THE DAMPED HARMONIC OSCILLATOR
84
Thus F = −kxi is a conservative force. Since F is a conservative force, we can find a scalar
potential V such that F = −∇V as follows
F = −kxi = −
This means that
∂V
= kx,
∂x
From the first equation, we obtain
∂V
∂V
∂V
i−
j−
k.
∂x
∂y
∂z
∂V
= 0,
∂x
∂V
= 0.
∂x
1
V = kx2 ,
2
which is the required potential energy of the simple harmonic oscillator.
Example 11.2. A particle of mass 5 moves along the x-axis under the influence of two forces
(i) a force of attraction to origin O which is numerically equal to 40 times the instantaneous
distance from O and (ii) a damping force proportional to the instantaneous speed such that
when the speed is 10 units/s, the damping force is 200 units. Assuming that the particle
starts from rest at distance 20 units from O.
(a) Set up the differential equation and initial conditions describing the motion.
(b) Find the position of the particle at any time t.
(c) Determine the amplitude, period and frequency of the damped oscillation.
(d) Sketch the graph of the motion.
Solution
(a) The force due to attraction is −40xi. For the damping force of magnitude f we have
f =β
dx
.
dt
When dx/dt = 10, f = 200, and so 200 = 10β or β = 20. Thus the damping force is
f = −20 dx
i. By Newton’s second law of motion we have
dt
5
or
dx
d2 x
i = −20 i − 40xi,
2
dt
dt
d2 x
dx
+ 4 + 8x = 0,
2
dt
dt
with initial conditions are x = 20, dx/dt = 0 at t = 0.
CHAPTER 11. THE DAMPED HARMONIC OSCILLATOR
85
(b) Let x = ert be the solution to the equation ẍ + 4ẋ + 8x = 0 then ẋ = rert , and ẍ = r2 ert .
Now, substituting for x, ẋ, ẍ and since ert ̸= 0 we have r2 + 4r + 8 = 0. So, r = −2 ± 2i
and thus the general solution is
x = e−2t (c1 cos 2t + c2 sin 2t).
At t = 0, x = 20 and dx/dt = 0 so we have c1 = 20 and c2 = 20. Hence the solution is
x = 20e−2t (cos 2t + sin 2t),
or
√
x = 20 2e−2t cos(2t − π/4).
√
(c) The amplitude is 20 2e−2t , the period is 2π/2 = π seconds, and the frequency is
vibrations per second.
1
π
(d) Figure
shows the graph of the motion, and in this diagram, the maximum amplitude
√ 11.2
−2t
is 20 2e .
√
Figure 11.2: Damped motion with maximum amplitude 20 2e−2t .
11.3
Exercises
1. A mass of 200 grammes placed at the lower and of a vertical spring stretches it 20cm.
When it is in equilibrium, the mass is hit and due to this goes a distance of 8cm before
coming down again. Find (a) the magnitude of the velocity imparted to the mass when
it is hit and (b) the period of motion.
CHAPTER 11. THE DAMPED HARMONIC OSCILLATOR
86
2. A vertical spring of constant k having natural length l is supported at a fixed point A.
A mass m is placed at the lower end of the spring, lifted up to a height h below A and
dropped. Prove that the lowest point reached will be at a distance below A given by
r
m2 g 2 2mgh
mg
+
.
l+
+
k
k2
k
3. Solve the differential equation
d2 x
dx
+ 2 + 5x = 0,
2
dt
dt
subject to the conditions x = 5, dx/dt = −3, at = 0, and give a physical interpretation
of the results.
4. A bead of mass m is constrained to move on a frictionless wire in the shape of a
cycloid whose parametric equations are x = a(ϕ − sin ϕ), y = a(1 − cos ϕ) which lies
in a vertical plane. If the bead starts from rest at point O, (a) find the speed at the
bottom of the path, and (b) show that the bead performs oscillations and find the
period of oscillation.
11.4
Answers
1. (a) At the equilibrium position of the object, the potential energy is zero whereas the
kinetic energy is given as K.E. = 12 mv 2 . After the particle has been hit, and has
gone down a distance of 8 cm or 0.08 m, the potential energy at that point of
stopping is P.E. = 12 kx2 = 12 k × (0.08)2 whereas the kinetic energy is zero. By
the principle of conservation of energy we have
1
1
× 200 × v 2 = k × (0.08)2
2
2
By Hooke’s law mg = kx, with m = 0.2 kg, g = 9.8 m/s2 , and x = 0.2 m. Thus,
k=
mg
0.2 × 9.8
=
= 9.8
x
0.2
Therefore, solving for v from
1
1
× 200 × v 2 = k × (0.08)2
2
2
will give v = 0.56 m/s or v = 56 cm/s.
(b) The period is given as T = 2π/ω, and ω 2 = k/m. Thus,
r
r
m
0.2
T = 2π
= 2π
= 0.9 seconds.
k
9.8
CHAPTER 11. THE DAMPED HARMONIC OSCILLATOR
87
2. If x is the stretch in the spring when a mass m is added then by Hooke’s law mg = kx or
x = mg/k. Let y be the distance of the lowest point of the particle from the equilibrium
position, and this will be taken as the reference point. At this point, kinetic energy
+ y)2 . At the
is zero, whereas the potential energy is P.E. = 12 k(x + y)2 = 21 k( mg
k
point where the mass is dropped, kinetic energy is zero whereas the potential energy
is P.E. = mg(x + y + h) = mg( mg
+ y + h). By the principle of conservation of energy
k
we have
2
mg
k mg
+ y = mg
+y+h
2 k
k
Solving for y gives
r
m2 g 2 2mgh
y=
+
k2
k
The lowest point will then be
r
mg
m2 g 2 2mgh
l+
+
+
k
k2
k
3. The auxiliary equation to this differential equation is r2 + 2r + 5 = 0 with two roots,
r1 = −1 + 2i and r2 = −1 − 2i. Thus the general solution is
x = e−t (c1 cos 2t + c2 sin 2t),
where c1 and c2 are arbitrary constants. Using the initial conditions x = 5, dx/dt = −3
at t = 0 we obtain c1 = 5 and c2 = 1, and therefore
x = e−t (5 cos 2t + sin 2t).
This may also be written as
√
26e−t cos(2t − ϕ),
√
√
√
where cos ϕ = 5/ 26 and sin ϕ = 1/ 26. The motion has amplitude 26e−t , period
of oscillation is T = 2π/2 = π, and frequency is f = 1/T = 1/π. The physical
interpretation of this is that the amplitude of oscillation decreases toward zero as t
increases
x=
4. (a) Let P be the position of the bead at any time t and let s be the arclength along
the cycloid measured from O. By conservation of energy, measuring potential
energy relative to the time AB through the minimum point of the cycloid, we
have the sum of potential and kinetic energy at P equals sum of potential and
kinetic energy at O, that is,
2
ds
1
= mg(2a) + 0
mg(2a − y) + m
2
dt
CHAPTER 11. THE DAMPED HARMONIC OSCILLATOR
88
Thus,
2
ds
dt
v =
2
= 2gy
At the lowest point, y = 2a, and the speed is
p
√
v = 4ga = 2 ag.
(b) Now, using
we know that
ds
dt
ds
dt
2
2
= 2gy,
=
dx
dt
2
+
dy
dt
2
,
where dx/dt = a(1 − cos α)ϕ̇ and dy/dt = a sin αϕ̇. So then
(a(1 − cos α)ϕ̇)2 + (a sin αϕ̇)2 = 2ga(1 − cos α).
This leads to ϕ̇2 = g/a or
r
ϕ=
g
t + c.
a
When ϕ = 0, t = 0, and when ϕ = 2π, t = T /2, where T is the period, and thus
using these initial conditions gives c = 0. Therefore
r
g
ϕ=
t.
a
p
p
This implies that 2π = ag T2 or T = 4π/ g/a and thus
r
T = 4π
a
.
g
Chapter 12
Forced vibrations
12.1
Introduction
Suppose that in addition to the restoring force−kxi and damping force −βvi we impress on
the mass m a force F (t)i where F (t) = F0 cos αt. Then the differential equation becomes
m
d2 x
dx
=
−kx
−
β
+ F0 cos αt
dt2
dt
or
ẍ + 2γ ẋ + ω 2 x = f0 cos αt,
(12.1)
where γ = β/2m, ω 2 = k/m, f0 = F0 /m. The general solution is obtained by adding the
complementary solution of ẍ + 2γ ẋ + ω 2 x = 0 to the particular solution of ẍ + 2γ ẋ + ω 2 x =
f0 cos αt. Now that the complementary solution is already obtained, we find the particular
solution by solving the equation ẍ + 2γ ẋ + ω 2 x = f0 cos αt using the method of undetermined
coefficients.
Now, let x = A cos αt + B sin αt be the particular solution. Then x = A cos αt + B sin αt
and ẋ = −αA sin αt + αB cos αt and ẍ = −α2 A cos αt − α2 B sin αt. Substituting these in
the given differential equation gives
−α2 A cos αt − α2 B sin αt + 2γ(−αA sin αt + αB cos αt) + ω 2 (A cos αt + B sin αt) = f0 cos αt
This simplifies to
(2γαB + (ω 2 − α2 )A) cos αt + ((ω 2 − α2 )B − 2γαA) sin αt = f0 cos αt
Equating the coefficients of cos αt and sin αt on both sides of the equation gives 2γαB +
(ω − α2 )A = f0 and (ω 2 − α2 )B − 2γαA = 0, and solving for A and B gives
2
A=
(ω 2 − α2 )f0
(ω 2 − α2 )2 + 4α2 γ 2
89
CHAPTER 12. FORCED VIBRATIONS
and
B=
90
2αγf0
(ω 2 − α2 )2 + 4α2 γ 2
Thus the particular solution is
(ω 2 − α2 )f0
2αγf0
cos αt + 2
sin αt
2
2
2
2
2
(ω − α ) + 4α γ
(ω − α2 )2 + 4α2 γ 2
f0
= p
(cos αt cos ϕ + sin αt sin ϕ)
(ω 2 − α2 )2 + 4α2 γ 2
f0
= p
cos(αt − ϕ),
2
(ω − α2 )2 + 4α2 γ 2
x =
(12.2)
where tan ϕ = 2αω/(ω 2 − α2 ). The general solution may now be written as a combination
of this and the solution to the homogeneous equation ẍ + 2γ ẋ + ω 2 x = 0. It should be noted
that solution to the homogeneous equation, called the transient solution, approaches zero
in a short time. After this time has elapsed, the motion of the mass m is now governed by
the particular solution, which is often called the steady-state solution. The vibrations which
take place are called forced vibrations and have a frequency equal to that of the impressed
force but lag behind by the phase angle ϕ.
12.2
Resonance
The steady-state solution of the forced vibration is
f0
cos(αt − ϕ)
x= p
2
2
(ω − α )2 + 4α2 γ 2 )
In this case the amplitude A is given by
f0
A= p
(ω 2 − α2 )2 + 4α2 γ 2
(12.3)
We assume that γ ̸= 0, i.e. β ̸= 0, so that the damping force is present. We shall prove
α
, of the impressed
that the maximum value of A in this case occurs when the frequency, 2π
1 2
2
2
2
force is such that α = αR = ω − 2γ assuming that γ < 2 ω . Near this frequency very
large oscillations may occur, sometimes causing damage to the system. This phenomenon is
called the resonance and the frequency α2πR is called the frequency of resonance or resonant
frequency. The value of maximum amplitude at the resonance frequency is
Amax =
f
p 0
2γ ω − γ 2
(12.4)
CHAPTER 12. FORCED VIBRATIONS
91
The amplitude may be written in terms of αR as
A=
(α2
−
2 2
αR
)
f0
+ 4γ 2 (ω 2 − γ 2 )
(12.5)
The graph of A against α2 is symmetric about the resonant frequency. In case there is
no damping, i.e. γ = 0 or β = 0, all the frequencies (resonant frequency, frequency with
damping, and natural frequency), are the same. In such case resonance occurs where the
frequency of the impressed force equals the natural frequency of oscillation. Then, the general
solution is,
f0 t
sin ωt
(12.6)
x = A cos ωt + B sin ωt +
2ω
From (12.6) it is seen that the oscillations build up with time until finally the spring
breaks.
12.3
Simple pendulum
A pendulum is a weight suspended from a pivot so that it can swing freely, as shown in
Figure 12.1. When a pendulum is displaced sideways from its resting equilibrium position, it
is subject to a restoring force due to gravity that will accelerate it back toward the equilibrium
position. When released, the restoring force combined with the pendulum’s mass causes it to
oscillate about the equilibrium position, swinging back and forth. The time for one complete
cycle, a left swing and a right swing, is called the period. The period depends on the length
of the pendulum, and also to a slight degree on the amplitude, the width of the pendulum’s
swing.
Here is a list of some of the assumptions when dealing with a pendulum.
1. Friction from both air resistance and the system is negligible.
2. The pendulum swings in a perfect plane.
3. The arm of the pendulum cannot bend or stretch/compress.
4. The arm is massless.
5. Gravity is a constant, 9.8 ms−2 .
Example 12.1. Prove that the amplitude of thep
steady-state oscillation is a maximum where
the resonant frequency is determined from α = ω 2 − 2γ 2 and that the amplitude is
f
p 0
2γ ω 2 − γ 2
CHAPTER 12. FORCED VIBRATIONS
92
Figure 12.1: A swinging pendulum.
Solution
The amplitude of the steady-state oscillation A is given by
f0
A= p
(ω 2 − α2 )2 + 4α2 γ 2
Since f0 is a constant, the amplitude is maximum when the denominator is minimum. So,
let u = (ω 2 − α2 )2 + 4α2 γ 2 . Now, the minimum value of u is obtained when du/dα = 0 and
d2 u/dα2 > 0, that is,
du
= −4α(ω 2 − α2 ) + 8αγ 2
dα
= α(2γ 2 − ω 2 + α2 ) = 0 or α2 = ω 2 − 2γ 2
d2 u
= −4(ω 2 − α2 ) + 8γ 2 = α2 + 8α2 (ω 2 − α2 )
dα2
= 8(ω 2 − 2γ 2 ) > 0.
p
Thus u is a minimum at α2 = ω 2 − 2γ 2 or α = ω 2 − 2γ 2 and the maximum amplitude
is
f0
Amax = p
(ω 2 − ω 2 + 2γ 2 )2 + 4γ 2 (ω 2 − 2γ 2 )
f0
= p
2γ ω 2 − γ 2
Example 12.2. Determine the motion of a simple pendulum of length l and mass m assuming small vibrations and no resistance force.
CHAPTER 12. FORCED VIBRATIONS
93
Solution
Let the position of m at any time t be determined by s, the arc length measured from the
equilibrium position O. Let θ be the angle made by the pendulum string with the vertical.
If e is the unit tangent vector to the circular path of the pendulum bob m, then the force
acting on it is −mg sin θe. By Newton’s second law of motion we have
m
d2 s
e = −mg sin θe
dt2
or
d2 s
= −g sin θ
dt2
Now the arclength is given by s = lθ so that equation (1) can be written as
d2 (lθ)
= −g sin θ
dt2
or
d2 θ g
+ sin θ = 0
dt2
l
If θ is small then sin θ ≈ θ, so that we obtain the linearised equation
d2 θ gθ
+
=0
dt2
l
(12.7)
(12.8)
Assuming the initial conditions θ = θ0 , dθ/dt = 0 at t = 0, the general solution is
r
r
g
g
θ = A cos
t + B sin
t.
l
l
Using the initial condition θ = θ0 at t = 0 we have θ0 = A cos 0 + B sin 0 and A = θ0 .
Thus,
r
r
g
g
θ = θ0 cos
t + B sin
t
l
l
and
dθ
=−
dt
r
g
θ0 sin
l
r
g
t+
l
r
g
B cos
l
r
g
t.
l
Now at t = 0, dθ/dt = 0, then
r
r
r
g
g
g
0=−
θ0 sin 0 +
B cos 0 =
B.
l
l
l
Since g/l ̸= 0, we have B = 0 and thus,
r
θ = θ0 cos
g
t.
l
CHAPTER 12. FORCED VIBRATIONS
94
The period is
2π
T = p g = 2π
s
l
l
,
g
and the frequency is given as
1
1
f= =
T
2π
r
g
.
l
Example 12.3. By using the principle of conservation of energy, show that
d2 θ g
+ sin θ = 0,
dt2
l
Solution
We see from figure above that OA = OC − AC = l(1 − cos θ). Then by the principle of
conservation of energy, the sum of the potential energy at B and kinetic energy at B equals
the total energy, E, that is,
2
1
ds
mgl(1 − cos θ) + m
=E
(12.9)
2
dt
since s = lθ, we have
2
1
d(lθ)
mgl(1 − cos θ) + m
=E
2
dt
or
2
1 2 dθ
mgl(1 − cos θ) + ml
=E
2
dt
Differentiating (12.10) with respect to time t we have
mgl sin θ
or
ml
2
dθ
d2 θ dθ
+ ml2 2
dt
dt dt
d2 θ g
+ sin θ
dt2
l
dθ
=0
dt
(12.10)
CHAPTER 12. FORCED VIBRATIONS
or
95
d2 θ g
+ sin θ = 0,
dt2
l
which is the required result.
12.4
Exercises
1. Find the solution if in addition to restoring force −kxi with no damping we impress
on the mass m a force F (t)i where F (t) = F0 cos αt and the impressed frequency of
oscillation, and give a physical interpretation.
2. A spring is stretched 5 cm by a force of 50 dynes. A mass of 10 grammes is placed on
the lower end of the spring. After equilibrium has been reached, the upper end of the
spring is moved up and down so that the external force acting on the mass is given by
F (t) = 20 cos ωt, t ≥ 0. Find (a) the position of the mass at any time, measured from
the equilibrium position, and (b) the value of ω for which resonance occurs.
3. A periodic external force acts on a 6 kg mass suspended from the lower end of a
vertical spring having constant 150 newtons/metre. The damping force is proportional
to the instantaneous speed of the mass and is 80 N when the speed is 2 ms−1 Find the
frequency at which resonance occurs.
4. Find the length of a simple pendulum whose period is 1 second. Such a pendulum
which registers seconds is called seconds pendulum.
5. A simple pendulum whose length is 2 metres has its bob drawn to one side until the
string makes an angle of 30o with the vertical and the bob is then released. Find the
(a) speed of the bob as it passes through the lowest point,
(b) angular speed at the lowest point, and
(c) maximum acceleration and where does it occur.
6. A simple pendulum of length and mass m hangs vertically from a fixed point O. The
bob is given an initial horizontal velocity of magnitude v0 . Prove that the arc through
which the bob swings in one period has a length given by
v02
−1
4l cos
1−
2gl
CHAPTER 12. FORCED VIBRATIONS
12.5
96
Answers
1. The restoring force is −kxi, the impressed force is (F0 cos αti, and therefore the effective
force is (−kx + F0 cos αt)i. By Newton’s second law of motion we have
m
or
d2 x
i = (−kx + F0 cos αt)i
dt2
d2 x
F0
k
cos αt = f0 cos αt
+ x=
2
dt
m
m
Since the impressed frequency is equal to the natural frequency of oscillation we have
α = ω, that is,
d2 x
+ α2 x = f0 cos αt
(12.11)
dt2
To find the general solution of this equation we add the solution of
d2 x
+ α2 x = 0
dt2
(12.12)
which is x = A cos αt+B sin αt, and called the complementary solution, to the solution
of ẍ + α2 x = f0 cos αt, called the particular solution. Now, to find the particular
solution, we assume the solution of the form x = t(c1 cos αt + c2 sin αt). However,
substituting this into the left side of (12.12) gives zero. Therefore we must modify the
form of the assumed particular solution to
x = t(c1 cos αt + c2 sin αt)
(12.13)
Differentiate (12.13) to obtain
ẋ =
dx
= t(−αc1 sin αt + αc2 cos αt) + c1 cos αt + c2 sin αt
dt
ẍ = t(−α2 c1 cos αt − α2 c2 sin αt) + 2(−αc1 sin αt + αc2 cos αt)
(12.14)
(12.15)
Substituting (12.13), (12.14) and (12.15) into (12.12), we find, after simplification
−2αc1 sin αt + 2αc2 cos αt = f0 cos αt
from which c1 = 0 and c2 =
f0
2α
Thus the required particular solution (??) is
x=
f0
t sin αt
2α
The general solution of (12.13) is therefore
x = A cos αt + B sin αt +
f0
t sin αt
2
(12.16)
CHAPTER 12. FORCED VIBRATIONS
97
The constants A and B in (12.16) are determined from the initial conditions. Unlike
the case with damping, the terms involving A and B do not become small with time.
However, the last term involving t increases with time to such an extent that the spring
will finally break.
2. By Hooke’s law, if 50 dynes stretches the spring by 5 cm, then using F = kx gives
50 = k × 5 or k = 10 dynes per centimeter. For the mass of 10 grammes, the force is
10 × 980 = 9800 dynes and so the extention of the spring will be x = 980 centimeters.
By Newton’s second law of motion we have
10
or
d2 z
k = −10zk + 20 cos ωtk
dt2
d2 z
+ z = 2 cos ωt
dt2
(12.17)
2
The solution of ddt2z + z = 0, called the complementary solution, is z = A cos t + B sin t.
Now, let z = −c1 cos ωt + c2 sin ωt be the solution of (1), called the particular solution.
Then
dz
= −c1 ω sin ωt + ωc2 cos ωt and
dt
d2 z
= −ω 2 c1 cos ωt − ω 2 c2 sin ωt
dt2
Substituting in the given equation we have
−ω 2 c1 cos ωt − ω 2 c2 sin ωt + c1 cos ωt + c2 sin ωt = 2 cos ωt
or
(1 − ω 2 )c1 cos ωt + (1 − ω 2 )c2 sin ωt = 2 cos ωt
Comparing coefficients of cos ωt and sin ωt on both sides gives c1 =
and therefore
2
z=
cos ωt
1 − ω2
Thus the general solution
z=
2
1−ω 2
and c2 = 0,
2
cos ωt + A cos t + B sin t
1 − ω2
Applying the initial conditions z = 980, dz
= 0, at t = 0 gives the constants A =
dt
2
980 − 2/(1 − ω ) and B = 0, and therefore
2
2
cos t
z=
cos ωt + 980 −
1 − ω2
1 − ω2
CHAPTER 12. FORCED VIBRATIONS
98
3. The force acting due to the tension in the spring is −150zk where z is the stretch in
the spring. The magnitude of the damping force is f = βv and when v = 2 metres per
second, f = 80 newtons. This gives 80 = 2β or β = 40 and so the damping force is
k. By Newton’s second law of motion we have
−40vk = −40 dz
dt
6
d2 z
dz
k = −150zk − 40 k
2
dt
dt
or
d2 z 20 dz
+ 25z = 0
+
dt2
3 dt
(12.18)
The auxiliary equation is
γ2 +
20
γ + 25 = 0
3
(12.19)
The solution of this equation is
γ=−
10 5 √
5i.
±
3
3
The general solution of the given equation is
−10
5√
5√
t
z=e 3
5t + B sin
5t
A cos
3
3
√
The angular velocity is ω = 5 3 5 and the period of oscillation is T =
the frequency at which resonance occurs is
√
5 5
6π
2π
ω
=
6π
√ .
5 5
Thus
4. The period T of a pendulum is given by
s
T = 2π
l
g
q
If T = 1 second then we have 1 = 2π gl or l = g/4π 2 = 24.60 cm. Thus the length of
a simple pendulum whose period is 1 second is 24.60 centimeters.
5. (a) We see from the figure that OA = OC − AC = 2 − 2 cos 30o = 2(1 − cos 30o ).
Then by the conservation of energy taking the reference level for potential energy
at the horizontal plane through the lowest point O we have the sum of potential
energy at B and kinetic energy at B as equal to sum of potential energy at O and
kinetic energy at O. So, now potential energy at B is myg 2 (1 − cos 30o ) whereas
kinetic energy at B is zero. Also, potential energy at O is zero whereas kinetic
CHAPTER 12. FORCED VIBRATIONS
99
energy at O is 12 mv 2 , where v is the speed of the bob at O. Thus by the principle
of conservation of energy we have
1
mg 2 (1 − cos 30o ) + 0 = 0 + mv 2
2
This means that v 2 = 4g(1 − cos30o ) = 2.3 meters per second, and this is the
speed of the bob as it passes its lowest point.
(b) We know that
ds
dt
= v and s = lθ. So then
d(lθ)
=v
dt
This implies that l dθ
= v or dθ
= vl . Since v = 2.3 and l = 2, we have dθ
= 1.15
dt
dt
dt
and therefore the angular speed at the lowest point is 1.15 radians per second.
(c) The equation of the pendulum is
g
sin θ = 0
l
θ̈ +
Thus magnitude of the angular acceleration is
θ̈ =
9.8
sin 30o = 2.45
2
radians per second, and acceleration = lθ̈ = 2 × 2.45 = 4.9 meters per second2 ,
which occurs at the angle 30o .
6. In one period the mass m moves from B to D and then back to B. Let s1 be the
distance along the arc from O to B. Then the required arc length is 4s1 . But s1 = lθ or
θ = s1 /l. Now, AO = CO − CA = l − l cos θ = l(1 − cos θ). Let the reference level for
potential energy be taken the horizontal plane through the lowest point O. The total
energy at O is the kinetic energy given by 12 mv02 . Now, since the velocity at B is zero,
the total energy at B is the potential energy mgl(1 − cos θ) = mgl(1 − cos(s1 /l)). By
the principle of conservation of energy we have
1 2
mv = mgl(1 − cos(s1 /l))
2 0
Solving for s1 gives
−1
s1 = l cos
v2
1− 0
2gl
and the required arc length is
−1
4s1 = 4l cos
v2
1− 0
2gl
Chapter 13
Elastic strings and springs
13.1
Introduction
When an elastic string is subjected to tensile forces it is found experimentally that the extension produced is proportional to the tension in the string. The same is also true of springs,
but in addition compressive forces will produce contractions for which the proportionality
law still holds. This law is known as Hooke’s law.
If l is the natural length of the string (or spring) and z is the extension produced by a
tensile force T , then
λz
T =
,
(13.1)
l
where λ is a constant known as the modulus of the string (or spring), which may also be
interpreted as the force required to double the length of the string (or spring) provided
Hooke’s law still holds for such a loading.
Essentially λ is a constant of the string (or spring) but it depends on the geometry as
well as the material of the string (or spring). In consequence it is sometimes useful to state
Hooke’s law in a form which involves quantities separately dependent on geometry and on
the material of the spring. Thus if we have a steel wire of natural length l, tensile force T ,
uniform cross sectional area A and extension z, then it is found that
z
T
=E
,
A
l
where E is a constant dependent only on the material and termed Young’s modulus. When
an elastic string is stretched, work is done by the tensile force, this work is stored as elastic
potential energy in the string. Thus if a tensile force T is applied to a string of natural length
l and modulus λ produces an extension z, the work done in producing a further extension
dz is
λz
dW =
dz,
l
100
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
so that the total energy stored in the string is
Z z λ
λz
dz = z 2 .
W =
l
2l
0
101
(13.2)
The form of work W indicates that elastic forces are conservative.
Consider now the motion of a particle P of mass m hanging from the lower end of an
elastic string of natural length l suspended vertically from a fixed point O. Let (l + x) be the
length OP at time t. The weight of the particle is mg and the tensile force in the string is
T =
λx
,
l
so that the net downward force is
λz
.
l
By Newton’s second law of motion, equation of motion is
mg −
mẍ = mg −
λz
.
l
(13.3)
The static equilibrium position is given by mg = λx/l or x = mgl/λ, and so putting
z = x − mgl/λ in (13.3) we have
λ
z̈ = − z.
(13.4)
ml
Equation (13.4) shows that so long as the string does not slacken, the particle executes
a rectilinear simple harmonic motion about the equilibrium position, the period being
ml
2π
λ
12
.
This requires that the depth z below the equilibrium position should never exceed mgl/λ or
that
2mgl
.
x≤
λ
Example 13.1. An elastic string of length 1.6 m and modulus of elasticity 30 N is stretched
between two horizontal points, A and B, which are a distance 2.4 m apart. A particle of
mass m is then attached to the midpoint of the string, and rests in equilibrium, 0.5 m below
the line AB. Find the value of m.
Solution. From the triangle ABC, it can be shown the the total length of the string
after the extension is 2.6 m. Therefore, for each half of the string, the natural length is 0.8
m, the extended length is 1.3 m, and the extension is (1.3 − 0.8) m = 0.5 m.
At the point of equilibrium,
2T sin θ = mg,
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
102
where T = λx/l, from Hooke’s law. Thus,
λx
2 sin θ = mg
l
30 × 0.5
0.5
2
×
= m × 9.81
0.8
1.3
which gives m = 1.47.
Example 13.2. Two light strings, S1 and S2 , are joined together at one end only. One end
of the combined string is attached to the ceiling at O, and a mass of 3 kg is attached to the
other, and allowed to hang freely in equilibrium. The moduli of S1 and S2 are 75 N and 120
N, and their natural lengths are 50 cm and 40 cm. Find the distance of the 3 kg mass below
O.
Solution.
We assume that the tension in the combined string is constant and that the masses of the
strings can be ignored. Let T be the tension in the combined string, x1 , x2 be the extensions
in the strings S1 , S2 , respectively.
The tension in an elastic string is given by T = λx/l, where λ is the modulus of elasticity,
x is the extension, and l is the natural length, of the string. For S1 , T = 3g, l = 0.5, and
therefore
75x1
,
3g =
0.5
which gives x1 = 0.1962. Similarly, for S2 ,
3g =
120x2
,
0.4
which gives x2 = 0.0981. Thus, the total distance of the 3g mass below O is 0.5 + 0.1962 +
0.4 + 0.0981 = 1.1943 metres.
Example 13.3. A box of weight 49 N is placed on top of a horizontal table. It is to be pulled
along by a light elastic string with natural length 15 cm and modulus of elasticity 50 N. The
coefficient of friction between the box and the surface of table is 0.4. If the acceleration of
the box is 20 cm s2 and the string is pulled horizontally, what is the length of the string?
Solution.
The tension in an elastic string is given by T = λx/l, where λ is the modulus of elasticity, x
is the extension, and l is the natural length, of the string.
Since the motion of the box is to the right, the resultant force is given by the equation
T − F = 5 × 0.2,
where the frictinal force F = µR = 0.4 × 49 = 19.6 and g has been taken as 9.8 ms−2 . Thus,
T = 19.6 + 1.0 = 20.6.
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
103
Figure 13.1: A box on top of horizontal table.
By Hooke’s law,
50 × x
= 20.6,
0.15
which gives x = 0.0618. Therefore, the extended length of the string is 0.15+0.0618 = 0.2118
metres.
T =
Example 13.4. Two elastic springs, S1 and S2 , are joined at each end, so that they are side
by side. The bottom end of the combined spring is placed on a table, and a weight of 60 N
is placed on the top. The moduli of elasticity of S1 and S2 are 80 N and 100 N, and their
natural lengths are 50 cm and 60 cm, respectively. Find the distance of the 60 N weight
above the table.
Solution.
Figure ?? shows the two cases, where the springs are at their natural lengths and when
they are compressed. The springs will have the same compressed length, but with different
compressions x1 and x2 .
In equilibrium position,
T1 + T2 = 60,
where
100x2
80x1
and T2 =
.
0.5
0.6
Since the compressed lengths of the springs are equal, then
T1 =
0.5 − x1 = 0.6 − x2
We now have two simultaneous equations, where we can solve for x1 and x2 , that is,
80x1 100x2
+
= 60
0.5
0.6
x1 − x2 = −0.1
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
104
Figure 13.2: Two springs carrying a mass on top.
This gives x2 = 0.2327 and therefore the weight of the mass above the surface of the
table is 0.6 − x2 = 0.3673 metres.
Example 13.5. An elastic spring, with natural length 30 cm and modulus of elasticity 42
N, is lying on a rough horizontal table, with one end fixed to the table at A. The spring is
held compressed so that the length of the spring is 24 cm. A teddy bear of mass 2 kg is
placed on the table at the other end of the spring, and the spring is released. If the friction
force is 5 N, find the speed of the teddy bear when the length of the spring is 29 cm.
Solution.
Figure 13.3 shows a compressed spring with one end fixed and a teddy bear at the other end.
The springs is initially compressed at a length of 24 cm and later at a compressed length of
29 cm, after it has been released.
The energy stored in an elastic spring is E = λx2 /2l, where λ is the modulus of elasticity,
x is the extension or compression, and l is the natural length, of the spring.
At a length of 24 cm, the extension is 0.06 m and thus
42 × 0.062
= 0.252,
2 × 0.3
and at a length of 29 cm, the extension is 0.01 m and thus
E=
42 × 0.012
= 0.007.
2 × 0.3
The energy released by the spring is 0.252 − 0.007 = 0.245 J, which increases the kinetic
energy. Again, work done by the frictional force of 5 N when teddy bear moves from a
distance of 24 cm to 29 cm is 5 × 0.04 = 0.20 J, which decreases the kinetic energy.
The final amount of kinetic energy is then given as the initial kinetic energy plus energy
released by the spring less work done by friction. That is,
E=
1
× 2v 2 = 0 + 0.245 − 0.20 = 0.045,
2
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
105
Figure 13.3: A compressed spring with one end fixed and a teddy bear at the other end.
where the v is the velocity of the teddy bear at a distance of 29 cm, and the initial kinetic
energy is taken to be zero. Thus v = 0.2121 implying that the speed of the teddy bear at a
distance of 29 cm is 0.2121 ms−1 .
Example 13.6. A climber is attached to a rope of length 50 m, which is fixed to a cliff face
at a point A, 40 metres below him. The modulus of elasticity of the rope is 9800 N, and the
mass of the climber is 80 kg. The ground is 80 m below the point, A, to which the rope is
fixed. The climber falls (oh dear!). Will he hit the ground?
Solution
Let x be the final extension of the rope. Now, the loss in potential energy is given as
E1 = mgh = 80g × (40 + 50 + x),
which increases the kinetic energy. The energy stored in an elastic spring is
E2 =
λx2
9800x2
=
.
2l
2 × 50
The final amount of kinetic energy is then given as the initial kinetic energy plus loss in
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
106
potential energy less elastic potential energy. That is,
9800x2
,
0 = 0 + 80g × (40 + 50 + x) −
2 × 50
or x2 − 8x − 720 = 0, which can be solved to get x = 31.1293. So, the climber would fall
121.1 metres if there is no ground, so he would hit the ground 120 mteres below, but at a
slower speed.
Example 13.7. An elastic string has natural length l, and is held vertically, with the upper
end fixed. A heavy particle, mass m attached to its lower end produces a static deflection d
the string if the particle is pulled down a further distance c below equilibrium and released,
show that in the case when d(d + 4l) > c2 > d2 , the time taken for the particle to return to
its original position of release is
12
d
(π − θ + tan θ),
2
g
where cos θ = d/c and θ is acute.
Solution
At the equilibrium position, the length of the string is l + d and we have
mg =
λd
,
l
or λ = mgl/d. When the string is extended a further distance x from the equilibrium
position, its length becomes l + d + x and the tension is
T =
mg
λd
(d + x) =
(d + x).
l
d
Since the weight of the particle is mg, the net downward force in this position is
mg
mg − T = −
x
d
and since the downward acceleration has magnitude ẍ, we have
mg
x,
mẍ = −
d
(13.5)
or ẍ = −gx/d. This equation holds only so long as the string is taut. The general solution
of (13.5) is
r
r
g
g
x = c1 cos
t + c2 sin
t
d
d
At t = 0, x = c so we have c1 = c. Also since ẋ = 0 at t = 0 we have c2 = 0 so solution of
(13.5) is
r
g
x = c cos
t.
d
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
107
This solution holds only as long as the string remains taut. But when x = −d the string has
attained its natural length. If t1 be the time taken for this to happen, then
r
g
t1 .
−d = c cos
d
p
p
Now, if cos θ = d/c, then − cos θ = c cos dg t1 or cos(π − θ) = c cos dg t1 . This means that
s
t1 =
d
(π − θ).
g
At this instant, the speed is given as
r
r
r
g
g
g
ẋ = −c
sin
t1 = −c
sin θ.
d
d
d
Thus at time t = t1 , the particle is rising against gravity with an upward velocity
r
g
v=c
sin θ.
d
Let t2 be the further time taken for the particle to attain its greatest height. The greatest
height the particle would attain under free gravitational flight would be
v2
c2
c2
d2
2
=
sin θ =
1− 2 ,
2g
2d
2d
c
since cos θ = d/c. If the string does not tighten in this phase of the motion, then the
maximum height should not exceed 2l, that is, c2 − d2 < 4ld or d(4l + d) > c2 . As this
condition holds, it follows that in the subsequent upward flight of the particle, the string
does not tighten. Hence
s
r
v
c g
d
t2 = =
sin θ =
tan θ
g
g d
g
and the total time from release to the highest point attained is
s
d
t1 + t2 =
(π − θ + tan θ)
g
Thus the total time required is
s
2
d
(π − θ + tan θ).
g
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
13.2
108
Exercises
1. Two points A and B in a horizontal plane distance l apart are joined by a light elastic
string of natural length l0 and modulus λ, where l0 < l. A particle of mass m is
attached to the mid-point of the string. This particle is drawn aside a small distance
and released. Show that the ensuing motion of the particle is approximately simple
harmonic and find the period.
2. A heavy particle of mass m hangs at the lower end of a light vertical elastic spring
of natural length l0 and modulus 2mg. Initially, the system is in equilibrium and the
upper end of the spring is made to execute vertical oscillations so that its downward
displacement at time t is a sin pt, where p2 = 2g/l0 . Find the displacement of the
particle at this instant.
13.3
Answers
1. The figure above shows the configuration at time t, and T is the tension in the strings
AP, BP. Then
21
1 2
2
l +z
AP = BP =
4
1
Therefore, the stretch of AP = 14 l2 + z 2 2 − 12 l0 ≈ 12 (l − l0 ), to first order. If z denotes
the displacement of the particle from O at time t, the equation of motion is
−2T cos θ = mz̈
Now,
cos θ =
z
=
AP
2z
,
1 ≈
1 2
l
l + z2 2
z
4
to the first order. Hence the equation of motion is
1
1
mz̈ = −4λ
−
z
l0
l
or
4λ(l − l0 )
z=0
ml0 l
and so the motion is simple harmonic with period given as
z̈ +
T = 2π
ml0 l
4λ(l − l0 )
12
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
109
2. Figure below shows the system in equilibrium at time t = 0. Then
l − l0
mg = 2mg
l0
so that l = 1.5l0 . In figure, the situation at time t is depicted, the particle having
displaced a distance x. Let T be the tension in the spring. The total stretch of the
spring is 0.5l0 + x − a sin pt and so
2mg 1
l0 + x − a sin pt
T =
l0
2
By Newton’s second law of motion we have mẍ = mg − T or
2mg 1
mẍ = mg −
l0 + x − a sin pt
l0
2
This may be written as
ẍ +
2g
2g
x = a sin pt,
l0
l0
(13.6)
or ẍ + p2 p2 a sin pt, where p2 = 2g/l0 . This equation exhibits the case of resonance and
its general solution is of the form
x = A cos pt + B sin pt
Now, let x = t(c1 cos pt + c2 sin pt) be a particular solution of (13.6). Then ẋ =
t(−pc1 sin +pc2 cos pt) + c1 cos pt + c2 sin pt) and ẍ = t(−p2 c1 cos pt − p2 c2 sin pt −
pc1 sin +pc2 cos pt − pc1 sin pt + pc2 cos pt = t(−p2 c1 cos pt − p2 c2 sin pt) − 2pc1 sin pt +
2pc2 cos pt. Substituting in the given equation we have
t(−p2 c1 cos pt−p2 c2 sin pt)−2pc1 sin pt+2pc2 cos pt+p2 t(c1 cos pt+c2 sin pt) = p2 a sin pt
Comparing the coefficients of sin pt and cos pt we obtain c1 = −ap/2 and c2 = 0. Thus
the particular solution is
ap
x = − t cos pt
2
CHAPTER 13. ELASTIC STRINGS AND SPRINGS
110
Hence the general solution is
x = A cos pt + B sin pt −
ap
t cos pt
2
Applying the initial conditions x = 0 and ẋ = 0 at t = 0 gives A = 0 and B = a/2 so
that
ap
a
x = sin pt − t cos pt
2
2
Chapter 14
Momentum and variable-mass
problems
14.1
Introduction
Consider an isolated system composed of two particles of masses m1 and m2 respectively.
An isolated system is a system on which no external forces act, Fext = 0; in such a system,
the total momentum is conserved. Internal forces are forces acted by one part of the system
on another, these are irrelevant for the motion of the system as a whole: they cannot change
the total momentum of the system. According to the Newton’s second law of motion,
dv1
dv2
= F21 and m2
= F12
(14.1)
dt
dt
A more general formulation, which allows to consider systems composed of several bodies,
or, equivalently a body with a varying mass, is the following:
m1
1. Define the momentum of particle i as pi = mi vi . The momentum is a vector. A system
composed of several particles has the momentum which is the vector sum of all the
constituent particle momenta:
X
X
P =
pi =
mi vi
(14.2)
i
i
2. Distinguish between internal forces and external forces. For the two particles acting
upon each other, by applying Newton’s third law of motion, the force acted by particle
1 on 2 is equal in size and opposite in direction to the one acted by particle 2 on 1.
This means that
dv2
dv1
= −m2
m1
dt
dt
or
d
dP
[m1 v1 + m2 v2 ] =
= 0,
dt
dt
namely the total momentum of an isolated system is conserved.
111
CHAPTER 14. MOMENTUM AND VARIABLE-MASS PROBLEMS
112
3. Assume now a non-vanishing net external force Fext on the system. By Newton’s second
law,
dP
= Fext
(14.3)
dt
namely, it states that the rate of change of the total momentum P of the system is
equal to the external force. This is a general formulation of Newtonian dynamics.
Momentum conservation of an isolated system can of course be obtained in the special
case of a vanishing external force. Clearly, another special case is the one of a body of
a fixed mass m, in which case the mass can be pulled out of the time derivarive on the
left hand side of (14.3), returning to the formulation of mẍ = F.
Newton’s second law in (14.3) is most conveniently used in its integrated form. Consider
for example the effect of the force between time t and time t + ∆t. Integrating (14.3) we get:
Z t+∆t
Z t+∆t
dP
Fext (t)dτ
dτ =
dt
t
t
The integral on the right hand side of this equation is called impulse. The statement of
Newton’s second law is then:
Z t+∆t
Fext (t)dτ
(14.4)
P (t + ∆t) − P (t) =
t
namely that the net change in the total momentum of the system over a period of time
equals to the impulse of the external force in this time. Among the examples of applications
in which the mass of the system is varying in time, are the rain drops falling vertically within
a cloud and the equation of motion of a rocket.
14.2
Mass accretion
Consider a system of mass m(t) and velocity v(t), which accretes, between time t and time
t + δt, an additional mass δm having velocity u(t). We assume that the time interval δt is
infinitesimally small. Let us write down the momentum before and after the accetion event
to be used on the left hand side of (14.4):
P (t) = m(t)v(t) + dmu(t)
P (t + dt) = m(t + dt)v(t + dt)
= m(t)v(t) + dt[m(t)v̇(t) + ṁ(t)v(t)] + O((δt)2 )
(14.5)
where in the second equation (corresponding to the momentum after the accretion) we have
expanded in δt and neglected terms of O((δt)2 )
Using now Newton law in (14.4) we compare the difference between the momenta before
and after the accretion event to the impulse of the external force:
(m(t)v(t) + δt[m(t)v̇(t) + ṁ(t)v(t)]) − (m(t)v(t) + δmu(t)) = F (t)δt
CHAPTER 14. MOMENTUM AND VARIABLE-MASS PROBLEMS
113
where we neglected O((δt)2 ) corrections. Clearly the non-infinitesimal term m(t)v(t) drops
in the difference on the left hand side. Dividing by δt and then taking the limit δt → 0
(where upon δm/δt → ṁ) we get:
m(t)v̇(t) + ṁ(t)(v(t) − u(t)) = F (t)
(14.6)
This is called the mass accretion equation.
14.3
The falling raindrop
Consider a raindrop falling vertically through a cloud, picking up water vapour as it descends.
For simplicity assume that the water vapour is stationary with respect to the ground (which
we use as our reference frame) so that u = 0. We further assume that the only external force
is the one due to gravity, and that it acts directly downwards:
F = −m(t)gk
where k is a unit vector pointing up. The problem is then one dimensional, with v =
Because we assumed u = 0, eq. (120) reduces to
dy
k.
dt
m(t)v(t) + ṁ(t)v(t) = −m(t)gk
which means that
d
[m(t)v(t)] = −m(t)gk
dt
or if we use a scalar variable for the velocity component downwards: v = −wk, we have:
d
(m(t)w(t)) = m(t)g
dt
(14.7)
This is the final result given what we assumed so far. Clearly, we have two unknown functions
of the time the mass m(t) and the velocity downwards w(t), which are coupled through the
mass accretion relation (14.7). Two functions cannot be fully determined by one constraint.
One clearly needs to further model the rate of accretion in order to solve the problem.
One simple model for the accretion rate is that the drops accumulate more vapour in
proportionality with their mass. If this is assumed one has,
ṁ(t) = αm(t)
where α is some constant. Under this assumption we can solve the problem. We obtain:
m(t)ẇ + αm(t)w(t) = m(t)g
so the mass drops and we get the equation:
ẇ + αw(t) = g
(14.8)
CHAPTER 14. MOMENTUM AND VARIABLE-MASS PROBLEMS
14.4
114
The rocket equation
This is another application of the momentum-impulse form of Newton’s second law, namely
the case of a rocket. For simplicity we shall consider the case of motion in space where one can
neglect external forces (gravity). Then (14.4) simply implies total momentum conservation:
P (t + ∆t) − P (t) = 0
(14.9)
The rocket acceleration is based on throwing the exhaust gas backwards at a high speed. The
motion is along the thrust axis, which we can assume, for simplicity to be constant in time.
The problem therefore reduces to one dimensional, and we can drop the vector notation (we
assume that the positive direction is forward, along the thrust axis).
The basic mechanism is depicted in figure above: a certain amount of gas of mass is
thrown backwards, with velocity u with respect to the rocket. We shall assume that this
velocity is constant throughout the flight. Thus, according to (14.9) the remaining body of
the rocket moves forward at a higher speed v + δv. Note that the total mass is conserved.
So if at time t the mass of the rocket was m(t), at (infinitesimally) later time t + δt the mass
can be obtained as a Taylor expansion:
m(t + δt) ≈ m(t) +
dm
δt + O((δt)2 ) = m(t) + dm
dt
The mass is decreasing, i.e. dm
< 0 so dm < 0. Thus, as shown in figure, the amount of
dt
mass thrown backwards is −dm > 0.
At time t, before throwing the amount −dm the momentum in some inertial frame is
P (t) = m(t)v(t)
After throwing it
P (t + ∆t) = (m(t) + dm)(v(t) + dv) + (−dm)(v + dv − u)
where the second term accounts for the momentum of the exhaust gas of mass −dm, which
moves with velocity −u with respect to the rocket, or v + dv − u with respect to the inertial
frame (where the rocket’s velocity is v + dv). Comparing the momenta before and after:
(m(t) + dm)(v(t) + dv) + (−dm)(v + dv − u) − m(t)v(t) = 0
CHAPTER 14. MOMENTUM AND VARIABLE-MASS PROBLEMS
115
and therefore, upon neglecting second order effects,
m(t)dv + udm = 0
Dividing by m(t) and integrating between the initial (i) and final (f ) times, we get:
Z mf
Z vf
dm
dv = −u
m
mi
vi
yielding:
mi
∆v = vf − vi = u ln
mf
which is the so-called Rocket Equation. It states that the difference between the final and
initial velocities is the exhaust velocity times the logarithm of the ratio between the initial
and final rocket mass.
Example 14.1. Derive an equation for the motion of a rocket moving in a straight line
under the influence of an external force of magnitude F .
Solution
Since the change in momentum of the system is equal to the impulse, we have total momentum at t + δt minus total momentum at t equals impulse, that is,
(m(t) + dm)(v(t) + dv) + (−dm)(v + dv − u) − m(t)v(t) = F dt,
This may be written as
dm
dv
+u
=F
(14.10)
dt
dt
Example 14.2. Find the velocity of the rocket of previous example assuming that the gas
is ejected at a constant rate and at constant velocity with respect to it and that it moves
vertically upward in a constant gravitational field.
Solution
If the gas is ejected at constant rate α > 0, then m = m0 − αt, where m0 is the mass of the
rocket at t = 0. Since F = −mg and dm/dt = −α, (14.10) can be written as
m
(m0 − αt)
or
dv
+ αu = −(m0 − αt)g
dt
dv
αu
= −g +
dt
(m0 − αt)
On integration, we get
v = −g + u ln(m0 − αt) + c1
(14.11)
At t = 0, v = 0 so that c1 = u ln m0 . Thus (14.11) becomes
m0
v = −g + u ln
m0 − αt
which is the speed at any time. Note that we must have m0 − αt > 0, otherwise there will
be no gas expelled from the rocket in which case the rocket will be out of fuel.
CHAPTER 14. MOMENTUM AND VARIABLE-MASS PROBLEMS
14.5
116
Exercises
1. A particle of varying mass m(t) is moving with velocity v under a force F(t), matter
being emitted at a constant rate λ with velocity u relative to the particle. Prove that
the total distance traveled by the rocket in time t is given as
1 2
m0 − αt
m0 − αt
s = − gt + v0 t + u
ln
2
α
m0
2. May be included later.
14.6
Answers
1. From (14.11) we have
m0
v = −g + u ln
m0 − αt
If s is the distance travelled by the rocket, then
m0
ds
= −g + u ln
dt
m0 − αt
Integrating gives
1 2 u
m0
s = − gt − (m0 − αt) ln
+1 +c
2
α
m0 − αt
But at t = 0, s = 0 so that c = m0 u/α and therefore
m0 − αt
1 2
m0 − αt
s = − gt + v0 t + u
ln
2
α
m0
Chapter 15
Motion under a central force
15.1
Introduction
Suppose that a force acting on a particle of mass m is such that (i) it is always directed
from mass m toward or away from a fixed point O, (ii) its magnitude depends only on the
distance from O. Then the force is called a central force of central force field with O as the
centre of force. In other words F is a central force if and only if it can be expressed as
r
F = f (r)r1 = f (r)
r
where r1 = r/r is a unit vector in the direction of r and r = |r| is the magnitude of r. The
central force field is an attraction toward O if f (r) < 0, otherwise a repulsion from O.
The properties of central force fields are
1. The path or orbit of the particle must be a place curve, i.e., a particle moves in a plane,
2. The angular momentum of the particle is conserved,
3. The particle moves in such a way that the position vector or radius vector drawn from
O to the particle sweeps out equal areas in equal times, i.e., the rate of change in area
is constant.
15.2
Equations of motion for a central force field
The motion of a particle in a central force field takes place in a plane. Choose this plane
to be the xy-plane and the coordinates of the particle as polar coordinates (r, θ). In polar
coordinates acceleration a is given by
a = (r̈ − rθ̇2 )r1 + (rθ̈ + 2ṙθ̇)θ1
(15.1)
Now, by Newton’s law of motion,
m(r̈ − rθ̇2 )r1 + (rθ̈ + 2ṙθ̇)θ1 = f (r)r1
117
(15.2)
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
118
Comparing coefficients of r1 and θ1 on both sides gives
m(r̈ − rθ̇2 ) = f (r),
(15.3)
m(rθ̈ + 2ṙθ̇) = 0.
(15.4)
and
From the (15.4) we have rθ̈ + 2ṙθ̇ = 0 or
d
(θ̇r2 ) = 0
dt
Integrating gives
θ̇r2 = h,
where h is a constant. This may also be written as θ̇ = h/r2 . Substituting in (15.3) we have
r−
h2
1
= f (r)
3
r
m
(15.5)
Now, let u = 1/r, then
du
du dr
1 dr
1 dr dt
=
=− 2
=− 2
dθ
dr dθ
r dθ
r dt dθ
Thus,
du
1 ṙ
ṙ
=− 2 =−
dθ
r θ̇
h
Differentiating gives
d2 u
1 d
1 d dt
=−
(ṙ) = −
(ṙ
2
dθ
h dθ
hdt dθ
r̈u
1
1
2 3
=−
= − 2 2 h u + f (r)
hu
m
hθ̇
1
= −u − 2 2 f (r)
hum
(15.6)
Again, we have
dr
dr dt
r2 ṙ
=
=
dθ
dt dθ
h
and
d2 r
1 d 2 1
r2
=
(r
ṙ)
=
(2rṙ2 + r2 r̈)
2
2
dθ
h dt
h
θ̇
4
3 2
2r ṙ
r r̈
=
+ 2
2
h
h 2r4 ṙ2 r4 f (r) h2
=
+ 2
+ 3
rh2
h
m
r
2
2 dr
r4
=
f (r)
+r+
r dθ
mh2
Equations (15.5), (15.6), and (15.7) are equations of motion in a central force field.
(15.7)
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
15.3
119
Potential energy in the central force field
A central force field is a conservative field, hence it can be derived from a potential. This
potential depends only on r and, apart from arbitrary constant, is given by
Z
V (r) = − f (r)dr
An arbitrary additive constant can be obtained by assuming, for example, V = 0 at r = 0
or V → 0 as r → ∞.
15.4
Determination of the orbit
If f (r) is given, it is possible to determine the orbit or path of the particle. This orbit can
be obtained in the form
r = r(θ)
that is, r is a function of θ or we can write
r = r(t) and θ = θ(t)
which are parametric equations in terms of the time parameter t.
Example 15.1. Prove that if a particle moves in a central force field, then its path must be
a plane curve.
Solution
Let F = f (r)r1 be the central force field. Then
r × F = r × f (r)r1 = f (r)r × r1 = 0
, this
since r1 is a unit vector in the direction of position vector r. Now that F = m dv
dt
equation can be written as
dv
r×m
=0
dt
This may also be written as
d
(r × mv) = 0
dt
Integrating this equation gives
r×v =h
where h is a constant vector. Multiplying through by r we obtain
r·r×v =r·h
Since r · r × v = r × r · v = 0 we have r · h = 0. Thus r is perpendicular to the constant
vector h and so the motion takes place in a plane. We shall assume that this plane is taken
to be the xy-plane whose origin is at the centre of force.
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
120
Example 15.2. Prove that the angular momentum of a particle moving in a central force
field is a constant.
Solution
We know that r · r × v = r · h, where h is a constant vector. Then multiplying this by mass
m we obtain
mr · r × v = mh
or r × mv = mh and since the left side is the angular momentum, it follows that the angular
momentum is a constant.
Example 15.3. Prove that a central force field is conservative and find the corresponding
potential energy of a particle in this field.
Solution
If we can find the potential V , then we will also have subsequently proved that the field is
conservative. Now if the potential V exists, it must be such that
F · dr = −dV
where F = f (r)r1 is the central force. We have
r
F · dr = f (r)r1 · dr = f (r) · dr = f (r)dr
r
since r · dr = rdr. Now we can determine V such that
−dV = f (r)dr
We have on integration
Z
V =−
f (r)dr
It follows that the field is conservative and that V represents the potential energy. In polar
coordinates, velocity v is given by
v = ṙr1 + rθ̇θ1
so that v 2 = v · v = ṙ2 + r2 θ̇2 . Thus the kinetic energy of the particle moving with velocity
v is given as
1
K.E. = (ṙ2 + r2 θ̇2 )
2
So the equation of conservation of energy is
Z
1 2
2 2
(ṙ + r θ̇ ) − f (r)dr = E
2
where E, a constant, is the total energy.
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
Now
121
dr
dr dθ
h dr
=
= 2
dt
dθ dt
r dθ
since θ̇ = h/r2 and therefore
" #
2
dr
+ r2 − f (r)dr = E
dθ
mh2
2r4
We may also write
m 2 h2
ṙ + 2 − f (r)dr = E
2
r
Now, let u = 1/r so that
dr
d 1
d 1 dθ
1 du
=
=
= − 2 θ̇
dt
dt u
dθ u dt
u dθ
Therefore
1 du 2
du
hu = −h
2
u dθ
dθ
" #
2
m 2 du
h2
h
+ 2 − f (r)dr = E
2
dθ
r
ṙ = −
and thus
or
1
mh2
2
This is then written as
"
#
2
du
+ u2 − f (r)dr = E
dθ
2
E−V
du
2
+u =2
.
dθ
mh2
Example 15.4. Show that the position of the particle as a function of time can be determined from the equations
Z
Z
1
− 21
t = (G(r)) dr, t =
r2 dθ
h
where
2E
2
G(r) =
+
m
m
Z
f (r)dr −
h2
2m2 r2
Solution
Consider the equation of energy
m
2
h2
ṙ + 2
r
2
− f (r)dr = E
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
122
which may also be written as
m 2
ṙ = E +
2
or
2
2E
+
ṙ =
m
m
Taking the positive square root gives
2
ṙ =
Z
mh2
f (r)dr −
2r2
Z
f (r)dr −
h2
= G(r)
r2
1
dr
= (G(r)) 2
dt
Integrating by the method of separation of variables gives
Z
1
t = (G(r))− 2 dr
If we write θ̇ = h/r2 or
dθ
dt
= h/r2 , then r2 dθ = hdt so that
Z
1
t=
r2 dθ
h
Example 15.5. Under the influence of a central force at point O, a particle moves in a
circular orbit which passes through O. Find the law of force.
Solution
In polar coordinates the equation of a circle of radius a passing through O is r = 2a cos θ.
Since u = 1/r, we can write u = sec θ/2a so that
sec θ tan θ
du
=
,
dθ
2a
Now
d2 u
sec3 θ + sec θ tan2 θ
=
dθ2
2a
u
1
d
2 2
+u
f
= −mh u
u
dθ2
3
sec θ + sec θ tan2 θ sec θ
2 2
= −mh u
+
2a
2a
3
2
sec θ + sec θ(tan θ + 1
2 2
= −mh u
2a
3 3
sec θ
sec θ
2 2
2 2 2
= −8mh u a
= −mh u
2a
a
2 2 5
= −8mh a u
−8mh2 a2
f (r) =
r5
Now since m, h, and a are constants, the force is one of attraction varying inversely as
the fifth power of the distance from O.
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
15.5
123
Exercises
1. Indicate which of the following central force fields are attractive toward origin O and
which are repulsive from O.
(a) F = −4r3 r1
(b) F =
√k r1 ,
r
(c) F =
r(r−1)
r
r2 +1 1
k>0
(d) F = (sin πr)r1
2. Prove that in rectangular coordinate system the magnitude of the areal velocity is gives
as
1
v = (xẏ − y ẋ).
2
3. Find the potential energy or potential corresponding to the central force fields defined
by
(a) F = − rk3 r1
(b) F = rα2 +
(c) F = krr1
(d) F =
β
r3
√1 r1 ,
r
k>0
r1
(e) F = (sin πr)r1
4. A particle moves in a central force field defined by F = −kr2 r1 . It starts from rest at
a point on the circle r = a. Prove that when it reaches the circle r = b its speed will
be
r
2π(a3 − b3 )
,
3m
and the speed will be independent of the path.
5. A particle of mass m moves in a force field
F=−
k
r1
rn
where k and r are constants. It starts from rest at r = a and arrives at r = 0 with
finite velocity v0 . Prove that we must have n < 1, k > 0, and
s
2ka1−n
v0 =
m(m − 1)
6. A particle moving in a central force field located at r = 0 describes the spiral r = e−θ .
Prove that the magnitude of the force is inversely proportional to r3 .
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
15.6
124
Answers
1. (a) Here f (r) = −4r3 , which is negative since r > 0 and thus the force is attractive.
(b) Since k > 0, the force is always positive so this is a repulsive force.
(c) This force is directed toward the origin, i.e., repulsive if 0 < r < 1, and repulsive
if r > 1.
(d) Here f (r) = sin πr. In this case, f (r) > 0 if 2nπ < πr < (2n + 1)π or 2n <
r < 2n + 1 for n = 0, 1, 2, ...and f (r) < 0 if (2n + 1)π < πr < (2n + 2)π or
2n + 1 < r < 2n + 2 for n = 0, 1, 2, ...
Thus F is repulsive for 2n < r < 2n + 1 and attractive for 2n + 1 < r < 2n + 2.
2. The areal velocity is given as
1
r×v
2
where r = xi + yj and where v = ẋi + ẏj. Thus the magnitude of areal velocity if
1
(xẏ − y ẋ).
2
3. The potential energy is given as
Z
V (r) = −
f (r)dr
√
(d) V (r) = 2 r
(a) V (r) = − 2rk2
(b) V (r) = αr + rβ3
(c) V (r) = − 12 kr2
(e) V (r) =
1
π
cos πr
4. The conservation of energy equation is
m 2
(ṙ + r2 θ̇2 ) −
2
Now,
R
f (r)dr =
R
Z
f (r)dr = E
kr2 dt = k3 r3 so that we have
m 2
k
(ṙ + r2 θ̇2 ) − r3 = E
2
3
At r = a, ṙ = 0, θ̇ = 0 since the particle starts from rest. Thus E = ka3 /3 and thus
m 2
k
k
(ṙ + r2 θ̇2 ) − r3 = a3
2
3
3
This may be written as
ṙ2 + r2 θ̇2 =
2k 3
(a − r3 )
3m
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
125
But the velocity is given as v 2 = ṙ2 + r2 θ̇2 and thus
v2 =
2k 3
(a − r3 )
3m
At r = b we have
2k 3
(a − b3 )
3m
which gives the required expression. Since the speed depends only on a and b, it is
independent of the path.
v2 =
5. Since f (r) = k/rn , the potential energy V is given by
Z
k
kr1−n
V (r) = −
+c
dr
=
rn
1−n
Since the potential is finite at r = 0 we must have 1 − n > 0 or n < 1. By the principle
of conservation of energy we have the sum of kinetic and potential energy equals E, a
constant. Thus
1 2 kr1−n
mv +
=E
2
1−n
At r = 0, v = v0 , and also at r = a we have v = 0 so that E = 12 mv02 and thus
v02 =
2ka1−n
m(n − 1)
Since n − 1 < 0 or n < 1 we must have
s
v0 =
2ka1−n
.
m(n − 1)
6. From Newton’s second law of motion we have ma = f (r)r1 , where f (r) is the magnitude
of force. In polar coordinates acceleration a is given by
a = (r̈ − rθ̇2 )r1 + (rθ̈ + 2ṙθ̇)θ1
where r1 is a unit vector in the direction of increasing r and θ1 is the unit vector in
the increasing θ-direction. Thus we obtain the two equations
m(r̈ − rθ̇2 ) = f (r)
m(rθ̈ + 2ṙθ̇) = 0
The second equation gives rθ̈ + 2ṙθ̇ = 0 or r2 θ̈ + 2rṙθ̇ = 0 after multiplying through
by r. This may be written as
d 2
(r θ̇) = 0
dt
CHAPTER 15. MOTION UNDER A CENTRAL FORCE
126
which is the same as r2 θ̇ = h, a constant. So we have θ̇ = h/r2 . Now from the first
equation,
h2
2
f (r) = m(r̈ − rθ̇ ) = m r̈ − 3
r
Now if r = r(θ) then
dr dθ
h dr
dr
=
= 2
dt
dθ dt
r dθ
d h dr
2h h dr
h2 d2 r
r̈ =
=
−
+
dt r2 dθ
r3 r2 dθ2
4 dθ2
ṙ =
and also
Equation (4) becomes
f (r) = m
h2 d2 r 2h2 dr
h2
− 3
− 5
4 dθ2
r
dθ2
r
If r = e−θ , then dr/dθ = −e−θ = −r and d2 r/dθ2 = e−θ = r so that equation (5)
becomes
2
2h2 2 h2
2mh2
h
r− 5 r − 3 =− 3
f (r) = m
4
r
r
r
Since m and h are constants, the magnitude of the force is inversely proportional to
r3 .