OSCILLATIONS AND SIMPLE HARMONIC MOTION
1.1
Definition of an Oscillating System
We begin our study of oscillations by examining the general definition of an oscillating system.
From this definition we can examine the special case of harmonic oscillation, and derive the
motion of a harmonic system.
An oscillating system is a system in which a particle or set of particles moves back and forth.
Oscillations occur when a system is disturbed from a position of stable equilibrium. This
displacement from equilibrium changes periodically over time. The back- end-forth motion is
repetitive and hence is said to be oscillatory and periodic, and display periodic motion.
Examples of oscillating system are a ball bouncing on a floor, a pendulum swinging back and
forth, a spring compressing and stretching, a balance wheel of a clock, the vibrations of strings
on musical instruments etc. The basic principle of oscillation maintains that an oscillating
particle returns to its initial state after a certain period of time. This kind of motion, characteristic
of oscillations, is called periodic motion, and is encountered in all areas of physics.
In every oscillating system there is an equilibrium point at which no net force acts on the
particle. A pendulum, for example, has its equilibrium position when it is hanging vertical, and
the gravitational force is counteracted by the tension. If displaced from this point, the pendulum
will experience a gravitational force that causes it to return to the equilibrium position. No matter
which way the pendulum is displaced from equilibrium, it will experience a force returning it to
the equilibrium point. If we denote our equilibrium point as x = 0, we can generalize this
principle for any oscillating system:
In an oscillating system, the force always acts in a direction opposite to the displacement of the
particle from the equilibrium point.
This force is called a restoring force. As long as the force obeys the above principle, the resulting
motion is oscillatory.
Many oscillating systems can be quite complex to describe. In our study we shall focus on a
special kind of oscillation called simple harmonic motion, which yields a simple physical
description.
1
In an oscillating system, the traditional variables x , v , t and a still apply to motion. But we are
going to introduce some new variables that describe the periodic nature of the motion: amplitude,
period, and frequency
1.2
Elastic Restoring Force
From Hooke’s law if an elastic string is pulled then the restoring force is related to the
displacement x from equilibrium position by
F = kx
1.3
Basic Concepts
Consider a body of mass m attached to
one end of a spring which is held
stationary at one end.
Let us describe the position of the body
with the coordinate x, with the origin
(x=0) at the equilibrium position.
When the body is displaced to the left the
spring is compressed and it exerts a force
on the body towards the right (the positive
x- direction) toward the equilibrium
When it is displaced to the right, x is positive, the spring exerts a force on the body toward the left
(the negative x-direction), toward the equilibrium position.
The sign of the x – component of the force on the body is always opposite to that of x itself.
Therefore if it obeys Hooke’s law
F = -kx ……………………………………………………………………….…………………(1)
where k is the constant of proportionality for the spring.
It is called the spring constant. k is a property of the spring.
Now suppose the body is displaced to the right a distance x = A, and released.
The spring exerts a restoring force; the body accelerates in the direction of the force and moves
toward the equilibrium position with increasing speed.
When the equilibrium position is passed the restoring force is directed to the right, the body’s
speed decreases and comes to rest at some point to the left of 0, and repeats its motion to the right.
The motion is confined to a range +A on either side of the equilibrium, with each to –and-fro
movement taking place in the same interval of time.
This type of motion, under the influence of an elastic restoring force proportional to displacement
and in the absence of all friction is called Simple Harmonic Motion (SHM).
2
Terminologies used in Oscillating Systems
It is important for us to understand the terminologies used in discussions on oscillating systems if
we are to be able to understand the system as a whole.
1. One round trip, say from A to –A and back to A, or from 0 to A to 0 to –A and back to A
to 0 is a complete cycle
2. The periodic time (T),
In simple oscillations, a particle completes a round trip in a certain period of time. This
time, T, which denotes the time it takes for an oscillating particle to return to its initial
position, is called the period of oscillation.
Therefore the periodic time (T), or simply called the period of the motion is the time
required for one complete cycle.
t
If a particle completes n round trips in a time t then T 
n
Period is measured in seconds.
3. We define frequency, denoted by f, as the number of cycles per unit time.
If a particle completes n round trips in a time t then f 
It can be easily seen that f 
n
t
1
T
Frequency is measured in Hertz (Hz), where 1 Hz = 1 cycle/second.
4. The amplitude, A, A simple oscillator generally goes back and forth between two
extreme points; the points of maximum displacement from the equilibrium point. We
shall denote this point by A, and define it as the amplitude of the oscillation.
Therefore the amplitude, A, is the maximum displacement from equilibrium, ie the
maximum value of x
If a body is displaced 1 cm from equilibrium and then allowed to oscillate we can say that
the amplitude of oscillation is 1 cm
The total range of the motion is 2A
5. Angular frequency or angular velocity is defined as the number of radians per second in
an oscillating system, and is denoted by ω.


t
where  is the angle covered in radians. One complete cycle =2π radians.
3
2.0
Simple Harmonic Motion
Having established the basics of oscillations, we now turn to the special case of simple
harmonic motion. We will describe the conditions of a simple harmonic oscillator,
derive its resultant motion, and finally derive the energy of such a system. Of all the
different types of oscillating systems, the simplest, mathematically speaking, is that of
harmonic oscillations.
2.1
Definition of Simple Harmonic Motion
Simple Harmonic Motion (SHM) is defined as oscillating motion of an object about a
fixed point such that the acceleration (a) of the object is:
1.
2.
proportional to the displacement (x) from the fixed point,
always directed towards the fixed point (the equilibrium position)
OR
Simple Harmonic Motion - is any motion that experiences a restoring force proportional
to the displacement of the system.
Consider a vibrating body of mass m, at some instant when its displacement from 0 is x.
The resultant force on it is F  kx
From Newton’s second law of motion the resultant force F  ma
Where a is the acceleration
Therefore  kx  ma
Therefore acceleration
k
a  ( ) x ………………………………...................……………(2)
m
When x has its maximum positive value A, the acceleration has its maximum negative value
k
a  ( ) A and vice versa.
m
When the particle passes the equilibrium position ( x  0 ), the acceleration is zero.
2.2
Energy of the system in SHM
The Potential energy U on the spring is
1
PE  kx 2 ………………………………………..…………………………….(3)
2
Kinetic energy of the particle
1
KE  mv 2 ………………………………………………..…………………….(4)
2
From the principle of conservation of energy the total energy E = KE + PE and is constant
4
Therefore
1 2 1 2
mv  kx = constant ……………………...…………………………….(5)
2
2
When the particle reaches it maximum displacement + A, it stops, therefore v  0
Thus total energy = Potential energy only
1
E  kA2
2
1
1
1
Thus total energy E = mv 2  kx 2  kA2 ………………………...……………………….(6)
2
2
2
This allows us to determine the velocity of the mass at any position x
v +
k
A2  x 2
m
………..…………………………………………………..………….(7)
At the midpoint x= 0, the energy is all kinetic and the speed is maximum, Vmax
1 2
mv max  E ,
2
…………………………………………………………….…………………….(8)
E
v max  2
m
But maximum potential energy = maximum kinetic energy = total energy
1
1
 kA2  mv 2 max ……………………………………………………………..……….(9)
2
2
Energy graph in SHM
2.3
Displacement in SHM
dx
Replace v by
from equation (7)
dt
dx
=
dt

5
dx
A2  x 2

k
A2  x 2
m
k
dt
m
x
k
sin 1 ( ) 
t c
A
m
𝑘
Therefore the displacement x = A sin (√𝑚 t + C ) ………………………………………(10)
where c is the integration constant. It determines the initial conditions of the motion, i.e. it
determines the position of the particle at time t = 0
2.4
Period and frequency of the motion
The period T of the SHM is determined when the quantity in parentheses () above increase by
2𝜋 from the time t = 0 to t = T
k
T  2
m
T  2
m
k
……………………………………………………………….………………..(11)
1
1
𝑘
Hence the frequency f =𝑇 = 2𝜋 √𝑚
f 
1
T
1
f 
2
……………………………………………….………(12)
k
m
The angular velocity  is defined as   2f
k
……………………………….……………………………..….… (13)
m
 is expressed in radians per second.

and hence equation (10) becomes x = A sin(  t +c) ……….……………………..…………(14)
Equation (7) becomes
……….……………………..…… (15)
v   A2  x 2
Recall: that the integration constant c determines the position of the particle at time t = 0
𝜋
𝜋
If at t = 0 c = 2 then x = A sin 2 = A
𝜋
In that case x = A sin (  t + 2 )
x =A cos  t
d 2x
Acceleration a  2   2 x ……….………………...…………………….…………...…… (16)
dt
6
Simple Pendulum
Another common example of objects moving with SHM is the simple pendulum. A classic
simple pendulum consists of a particle suspended from a light cord. When the particle is pulled
slightly to a small angle  and released, it swings back past the equilibrium point and oscillates
between two maximum angular displacements with periodic motion. We want to establish that it
executes simple harmonic motion.
We do so by drawing a free body diagram and examining the forces on the pendulum at any
given time.
Figure: A simple pendulum with cord of length L, shown with free body diagram at a
displacement of θ from the equilibrium point
The pendulum has been displaced through an arc of a circle whose centre is at the suspension
point and has length s.
The two forces acting on the pendulum at any given time are tension T from the cord and gravity
mg. At the equilibrium point the two are antiparallel and cancel exactly, satisfying our condition
that there must be no net force at the equilibrium point. When the pendulum is displaced by an
angle θ, the gravitational force must be resolved into radial and tangential components. The
radial component, mg cosθ, cancels with the tension, leaving net tangential force
F= - mg sinθ
The restoring force F  mg sin  .
Since this must obey Newton’s law of motion, F  ma
Then ma  mg sin  .
7
Therefore the acceleration of the mass a   g sin 
Mathematically, when the angle  is small and is expressed in radians
Therefore we can write
a   g .
But  
sin   
s
l
s
Therefore a   g .
l
g
 a  ( ) s
l
g
Note that for a given pendulum g and l are constant, hence ( )  constant. Therefore the
l
acceleration is proportional to the displacement s from the equilibrium position. Hence the
motion is SHM.
g
Compare a  ( ) s with a   2 x
l
g
Therefore ( )   2
l
2
2

The period of the motion T 

g
l
Therefore a simple pendulum executes SHM with period T  2
l
g
Note:
the period, and the frequency, of the pendulum is independent of the mass of the particle
 on the cord. It only depends on the length of the pendulum and the gravitational constant.
this is only an approximation. If the angle exceeds more than 15 degrees, the
 approximation breaks down
8
Example
1. A spring is mounted is fixed to a rigid support.
By attaching a spring balance to the free end and pulling sideways, it is found that the force is
proportional to the displacement, a force of 4N causing a displacement of 0.02m.
A body of mass 2kg is then attached to the end, pulled aside a distance of 0.04m, and
released
(a) Find the spring constant
k
Solution
F
4N

 200 Nm 1
x 0.02m
(b) Find the period and frequency of the vibration
Solution
T  2
m
2kg
 2
k
200 Nm 1
= 0.628 s
Frequency
f 
1
1

T 0.628
= 1.59 Hz
Angular frequency
  2f 
k
m
= 10s-1
(c) Compute the maximum velocity attained by the vibrating body.
The maximum velocity occurs at the equilibrium position where
From
x0
v   A2  x 2
When
x0
Vmax = +(10s-1)(0.04m) = + 0.4ms-1
(d) Compute the maximum acceleration
a
k
x
m
 2x
The maximum acceleration occurs at the ends of the path, where x =  A
Therefore,
amax    x
= + (10s-1)2(0.04m)
= + 4.0 ms-2
(e) Compute the velocity and acceleration when the body has moved halfway in toward the
center from its initial position
2
9
At this point
x
 0.02m
A 0.04

2
2
v  (10s 1 ) (0.04m) 2  (0.02m) 2
 0.346ms 1
Acceleration
a   2 x
 (10ms 1 ) 2 (0.02m)
 2.0ms 2
(f) How much time is required for the body to move halfway in to the center from its initial
position?
The position at any time is given by
x  A cos t
when halfway in to the center
Therefore
x
A
2
A
 A cos t
2
cos t 
1
2
1
cos 1 ( )  10t
2

3
 10t
t

30
s
(g) Find the maximum potential energy of the system
Potential energy
P.E 
1 2
kx
2
Maximum PE is obtained when
Therefore PEmax
10
P.Emax 
xA
1 2
kA
2
1
 x 200 x0.04 2
2
 0.16 J