4.1 Kinematics of simple harmonic motion (SHM)
Terminology:
Hooke’s law gives the force exerted by a spring that is stretched or compressed beyond the
equilibrium position. The force is proportional in magnitude to the displacement from the
equilibrium position.
Simple Harmonic Motion as exhibited by the mass spring system is one form of periodic
motion. The spring force is a restoring force always directed toward the equilibrium position.
The Acceleration of an object in simple harmonic motion (SHM) is proportional to the
displacement from equilibrium and oppositely directed. a=-(k/m)x is the result of applying
Newton’s second law to a mass (m) where the force is given by f= -kx. Note that the
acceleration of a harmonic oscillator is not constant.
For an ideal oscillator in harmonic motion along the x-axis, the following constants of motion
are found in expressions for position, velocity, and acceleration.
Amplitude, A, represents the maximum displacement along either positive or negative
x.
Angular frequency, ω, has units of radians per second and is a measure of how
rapidly the oscillations are occurring. The frequency f, measured in hertz (Hz),
represents the number of oscillations per second. Frequency and angular frequency
are related as stated in the equation ω =2πf.
Phase Constant, Ф, is determined by the position and velocity of the oscillating
particle when t=0. The quantity (ωt + Ф) is called the phase.
The Mathematical representation of a mass in simple harmonic motion along the xaxis is a second order differential equation. The ratio k/m is denoted by the factor ω2.
(d2x/dt2)= -ω2x (See graph and explanation below)
x=A cos(ωt + Ф)
2πs
Ф
Displacement-time
graph for an object
undergoing SHM
For the simple harmonic motion shown in the partial graph above:
A=4 units of length
ω =1 rev/ 2πs = 2π rad/ 2πs =1 rad/ s
Ф = +( π/2) rad
Description and problems:
In mechanical systems, energy can be stored in springs. For example a compressed spring on
a popgun can provide energy to a pellet and shoot it from the gun. Even a simple slingshot
uses the elastic energy stored in a rubber band to shoot it projectile. When two balls collide it
is as if a spring is between the balls. During the collision, the spring is compressed and
eventually the spring throws the balls apart again. In a case such as this, the spring is very
complex. Bonds within the balls’ molecules are distorted or deformed. F=kx is the equation
for Hooke’s law. Hooke’s law simply states that the magnitude of the distortion is directly
proportional to the magnitude of the distorting force, and their directions are the same. k is
the spring constant. The external deforming force is proportional to the deformation, which is
there represented as x.
All of us are familiar with the vibration of a mass at the end of a spring. The general behavior
is seen in the figure below. The mass at the end of the spring leaves a record on the paper,
which shows how the mass oscillates up and down. (This harmonic oscillation is how a
seismograph works to measure the harmonic tremors in the ground before a volcanic
eruption.) On complete vibration or cycle of the mass occurs when it vibrates from the
position indicated by point A to the point indicated by C or by any two other similar points.
This type of motion is denoted as a periodic or vibratory motion. The time taken for the
system to undergo on complete vibration is called T. Since the system will undergo 1/T
complete vibrations in unit time, this quantity is called the frequency of the vibrations (f or v)
and the formula will be f=1/T.
Frequency is usually expressed as cycles or vibrations per second. We denote the distance
AD as the amplitude of the vibrations. The dashed line notes the equilibrium position.
Vibratory motion of a system, which obeys Hooke’s law, is called Simple Harmonic Motion.
Problem 1:
The position of a particle is given by the expression x= (4.00m) cos(3.00πt + π), where x is in
meters and t is in seconds. Determine (a) the frequency and period of the motion, (b) the
amplitude of the motion, (c) the phase constant, and (d) the position of the particle at
t=0.250s.
(I am going to walk you through this one and then you will need to try the next one before
getting the answer off line…)
1. The given expression describes the vibration by telling the position at every instant of
time as it continually changes. Think about a treetop swaying in the wind while you
are working on this. The frequency, period, amplitude, and phase constant are
constants that describe the whole motion…
2. This is an x versus t function. We use the particle in simple harmonic motion model.
3. The particular position function x= (4.00m) cos(3.00πt + π) and the general one from
the graph above x=A cos(ωt + Ф) have functional equality. They must have the same
x value for all values of the variable t. This requires that everything be the same.
4. Let’s start with the formulas.
a. ω = 3.00π rad/s = (2π rad/1 cycle) f
3.00π rad/s
=f
2π rad/1 cycle
1.50Hz = f
T=1/ f
=1/1.5= 0.667s
b. A is the amplitude or distance in meters that a way moves.
A= 4.00m
c. Ф= π rad
d. At t= 0.250s,
x= (4.0m)cos (2π x1.5cycle x0.25s + π)= (4.0m)cos(5.50)
x= 4.0m x (0.709)
x= 2.83m
5. This vibration would be impressive to watch. The top of the tree would swing back
and forth through a range of eight meters, making 15 complete vibrations every ten
seconds. (Tornado weather)
6. Make sure you recognize the difference between Hz and radians per second.
Problem 2:
A particle moving along the x-axis in simple harmonic motion starts from its equilibrium
position, the origin at t=0 and moves to the right. The amplitude of its motion is 2.00cm and
the frequency is 1.50 Hz. (a) Show that the position of the particle is given by
x=(2.00cm)sin(3Πt), (b) determine the maximum speed and the earliest time (t>0) at which
the particle has this speed, (c) the maximum acceleration and the earliest time (t>0) at which
the particle has this acceleration and (d) the total distance traveled between t=0 and t=1.00s.
Below you will find the analysis, but before we begin that, think about the problem this
way… in its period of less than a second the particle moves 8 cm. Its minimum speed is 0 so
its maximum speed must be several cm/s and its maximum acceleration a larger number of
cm/s2.
Analysis:
(a) At t=0, x=0 and v is positive. The sine function is zero and the cosine is positive at
Ф=0, so this situation responds to x=Asin ωt and v=vi cos ωt.
f = 1.50 Hz
A= 2.00cm
ω=2π f or 3 π/ s
Therefore, x=(2.00cm)sin(3πs-1t)
(b) This is equal to writing x=A cos(ωt + Ф)
With
A= 2.00cm, w=(3πs-1) and Ф=π/2
Note that t=1/ f or 1/1.5=t , t=0.667s
To find velocity, Δx/Δt = v
V= 2.00(3.00π) cos(3.00πt) cm/s
Vmax= 2.00 x 3.00π when cos(3.00πt)=+1
Vmax= 18.8cm/s when t=0
And next at t=T/2 or 0.667/2 or 0.333s when cos[(3.00πs-1) (0.333s)]=-1
(c) Again a=Δv/Δt= (-2.00cm) (3.00πs-1)2 sin (3.00πt)
amax= A ω2 = (2.00cm) (3.00πs-1)2= 178cm/s2
(d) A=2cm then at 1 period the particle would travel distance =8cm
.667=2/3T thus at 1.00 sec the particle could travel
3/2(8.00cm)=12.0cm
IB Information
4.1.1 Describe examples of oscillations.
Car suspensions
Vibrations of the strings of a musical instrument
The swinging of a pendulum in a clock
4.1.2 Define the terms displacement, amplitude, frequency, period and phase difference.
The connection between frequency and period should be known.
4.1.3 Define simple harmonic motion (SHM) and state the defining equation as a = -ω2x.
Students are expected to understand the significance of the negative sign in the
equation and to recall the connection between ω and T.
4.1.4 Solve problems using the defining equation for SHM.
(See equations above)
4.1.5 Apply the equations listed above as well as:
x = x0 cos ωt and x = x0 sin ωt as solutions to the defining equation for SHM.
V= ±ω√x02-x2
4.1.6 Solve problems, both graphically and by calculation, for acceleration, velocity and
displacement during SHM.
More Problems will be attached before the weekend in a separate sheet. This hopefully will
help you start this section.