This document discusses simple harmonic motion (SHM). SHM occurs when an object experiences a restoring force proportional to its displacement from equilibrium. This results in sinusoidal oscillations described by x(t) = Acos(ωt + φ), where A is amplitude, ω is angular frequency, and φ is the phase. SHM includes examples like a mass on a spring and a simple pendulum. The relationships between displacement, velocity, acceleration, period, frequency, and energy in SHM systems are explored.

1. Wave phenomena
Topic 9.1 Simple Harmonic
Motion

2. Simple Harmonic Motion
In order for an object to perform simple
harmonic motion there must be a restoring
force. This leads to :
a)Magnitude of the force (and therefore the
acceleration) is proportional to the
displacement from the equilibrium point.
b)The direction of the force (and therefore the
acceleration) is always towards the equilibrium
point.

3. This leads to the mathematical expression
a α -x
And hence
a = -kx
Which graphically looks like:
a
x

4. Displacement
Its position x as a function of time t is: where A is the
amplitude of motion: the distance from the centre of motion
to either extreme
Its position x as a function of time t
O
B AP
( )tAx ωcos=

5. Angular frequency
Angular frequency is the rotational analogy to frequency. Represented
as , and is the rate of change of an angle when something is moving in a
circular orbit. This is the usual frequency (measured in cycles per
second), converted to radians per second. That is
f
T
π
π
ω 2
2
==
Which ball has the larger angular frequency?
Ball B

6. SHM and circular motion
Uniform Circular Motion
(radius A, angular velocity w)
Simple Harmonic Motion
(amplitude A, angular frequency w)
Simple harmonic motion can be visualized as
the projection of uniform circular motion onto one
axis
The phase angle ωt in SHM corresponds to the
real angle ωt through which the ball has moved in
circular motion.

7. Velocity and acceleration
Once you know the position of the oscillator for all times, you can work out
the velocity and acceleration functions.
x(t) = A cos (ωt + φ)
The velocity is the time derivative of the position so:
v(t) = -A ω sin (ωt + φ)
The Change from cos to sin means that the velocity is 90o
out of
phase with the displacement
when x = 0 the velocity is a maximum and when x is a minimum v = 0
The acceleration is the time derivative of the velocity so:
a(t) = -A ω2
cos (ωt + φ)
Notice also from the preceding that: a(t) = -ω2
x
The acceleration is exactly out of phase with the displacement.

8. Questions
1. A pendulum has maximum horizontal displacement
of 0.10m. It oscillates with a period of 2s.
• What is it’s displacement at time 0.5s
• What is it’s displacement at time 1.3s
• What is the maximum velocity
• What is the maximum acceleration
2. A surfer bobs up and down on the surface of a
wave with a period of 4.0s and an amplitude of
1.5m.
• What is the surfer’s maximum acceleration
• What is the surfer’s maximum velocity

9. Questions
3. An object moving with SHM has an amplitude of
2cm and a frequency of 20 Hz
– What is it’s period
– What is its acceleration at the middle and end of an
oscillation
– What are the velocities at the middle and end
4. A steel strip clamped at one end, oscillates with
frequency of 50Hz and has an amplitude of 8mm
– What is its period
– What is its angular frequency
– What is its velocity at the middle and end of an oscillation
– What are the corresponding accelerations

10. Summary 2
There are many relations among these parameters. One
minimum set of parameters to completely specify the motion
is:
amplitude (A)
angular frequency (ω)
initial phase (φ)
You are already familiar with this set, which is used in:
x(t) = A cos (ωt + φ)
The trick in solving SHM problems is to take the given
information, and use it to extract A, ω and φ. Once you have A,
ω and φ, you can calculate anything.

11. Summary 3
For a given body with SHM
( )txx ωsin0=
( )txv ωω cos0= xxv −= 2
0ω
( )txa ωω sin2
0−= xa 2
ω−=
In terms of time In terms of
displacement
Displacement
Velocity
Acceleration

12. Graphing shm
http://www.physics.usyd.edu.au

13. There are 2 practical examples of SHM
•Mass on a vertical spring
•The simple pendulum
Mass on vertical spring
Uses Hooke’s Law F = kx
And F=ma
To give T = 2 π√(m/k)

14. Simple pendulum
There is a component of the weight of the bob
acting towards the centre of the motion
We can use Newton's 1st
and 2nd
law to help us
T + Wcosθ =0 perpendicular
and
Wsin θ = -ma horizontally
W
T
θ

15. But W = mg
So,
mg sin θ = -ma
g sin θ = -a
If sin θ = θ then, θ = x/l
g x / l = -a
But we know for SHM
a = - kx
So, k = g/l
This results in a relationship with the period as
T = 2π√(l/g)

16. We can use this equation to establish a value for g
Practical:
•Set up simple pendulum
•Decide on a set point in the oscillation for counting
•Measure the time taken for about 20 oscillations for
5 lengths up to 50cm
•Calculate period
•Plot length against period squared
•Use gradient to find g….gradient = 4 π2
/g

17. • Animated video of shm
• Visit physclips

18. The frequency of simple harmonic motion like a mass on a spring is
determined by the mass m and the stiffness of the spring expressed in
terms of a spring constant k ( see Hooke’s Law):

19. Mass on spring resonance
A mass on a spring has a single resonant frequency
determined by its spring constant k and the mass m. Using
Hooke’s law and neglecting damping and the mass of the
spring, Newton’s second law gives the equation of motion:
the expression for the resonant vibrational
frequency:
This kind of motion is called simple
harmonic motion and the system a
simple harmonic oscillator.

20. Mass on spring: motion sequence
A mass on a spring will trace out a sinusoidal pattern as a
function of time, as will any object vibrating in simple harmonic
motion. One way to visualize this pattern is to walk in a straight
line at constant speed while carrying the vibrating mass. Then
the mass will trace out a sinusoidal path in space as well as
time.

21. Energy in mass on spring
The simple harmonic motion of a mass on a spring is an
example of an energy transformation between potential energy
and kinetic energy. In the example below, it is assumed that 2
joules of work has been done to set the mass in motion.

22. Potential energy
22
2
0
0
2
1
2
1
xm
kx
kxdx
FdxE
x
x
p
ω=
=
=
=
∫
∫At extension x:






=
m
k2
ω

23. Kinetic energy
At extension x:
( )22
0
22
2
1
2
1
xxmmvEk −== ω

24. Total energy
Total energy
( )
2
0
2
2
0
222
2
1
2
1
2
1
xm
xxmxm
EE kp
ω
ωω
=
−+=
+=