Simple Harmonic Motion

1
Simple Harmonic Motion

• Any motion which repeats its in equal intervals
  of time is termed periodic motion.

• If the motion is back and forth over the same
  path, then it is termed oscillatory or harmonic.

• Simple harmonic motion (SHM) deals with motion
  which oscillates without loss of energy due to
  friction, air resistance etc.

2
Simple Harmonic Motion

• Consider a body oscillating in the vertical axis
  between 5 and -5.

The displacement of the object as a function of
time is a sine wave
3
Simple Harmonic Motion

• The displacement can be expressed as x Acos(wt
  f)

A amplitude
t
w angular frequency 2pf
f frequency, number of complete oscillations per
unit time
T time taken for one oscillation. T 1/f
f phase angle (radians)
4
Simple Harmonic Motion

• The displacement can be expressed as x Acos(wt
  f)

t
The velocity is v dx/dt -wAsin(wt f)
??A2 - x2
The acceleration is a dv/dt -w2Acos(wt f)
-w2x
5
Simple Harmonic Motion

• Example The equation of motion of a simple
  harmonic oscillator is
• x 4cos(pt p/4)
• 1. Determine the amplitude, frequency and period.

Comparing with the general equation x Acos(wt
f)
Then
amplitude is A 4 m
angular frequency is p
?
frequency is w/2p 0.5 Hz
period is 1/f 1/0.5 2 s
6
Simple Harmonic Motion

• Example The equation of motion of a simple
  harmonic oscillator is
• x 4cos(pt p/4)
• 2. Determine the position, velocity and
  acceleration at time t 1 s.

Position is x 4cos(p p/4)
4cos(5p/4)
4?(-0.707)
-2.83 m
Velocity is v -4psin(p p/4)
-4psin(5p/4)
-4p?(-0.707)
8.88 m/s
Acceleration is a -w2x
-p2(-2.83)
27.9 m/s2
7
Simple Harmonic Motion

• Example The equation of motion of a simple
  harmonic oscillator is
• x 4cos(pt p/4)
• 3. What is the maximum and minimum values of the
  velocity and acceleration?

Velocity is x -4psin(pt p/4)
Max velocity occurs when sin(pt p/4) 1
vmax 4p m/s
This occurs at x 0 m
Acceleration is a -w2x
4p2 m/s2
Max acceleration is amax w2xmax
w2A
This occurs when x is a maximum
8
Examples of SHM-Mass of a Spring

• Release the mass connected to a spring and the
  spring will exert a restoring force with
  magnitude

F-kx
where k is the spring constant
But F ma
? -kxma
?
a -(k/m)x
Comparing with a -w2x
then
w2 k/m
? The body exhibits SHM
9
Examples of SHM-Mass of a Spring

• Example A light spiral spring has a mass of 25 g
  attached to its end. An extra 0.5 g stretches
  the spring 4.9 cm. This extra mass is removed
  and the mass is set vibrating with an amplitude
  of 10 cm. Find
• (a) The period of the vibration

First find the spring constant.
k Displacement force/displacement (0.5 x
10-3 x 9.8)/4.9 x 10-2 0.1 Nm-1
10
Examples of SHM-Mass of a Spring

• Example A light spiral spring has a mass of 25 g
  attached to its end. An extra 0.5 g stretches
  the spring 4.9 cm. This extra mass is removed
  and the mass is set vibrating with an amplitude
  of 10 cm. Find
• (b) The velocity and acceleration of the mass
  when its displacement is 6 cm from its mean
  position

v ??A2 - x2
a -?2x
(? 2?/T 2 rad.s-1)
-(2)2 x 6 x 10-2
2 ?(10-1)2 - (6 x 10-2)2
-24 x 10-2 ms-1
2 ?100 x 10-4 - 36 x 10-4
2 ?64 x 10-4
2 x 8 x 10-2
16 x 10-2 ms-1
11
Examples of SHM-Mass of a Spring

• Example A light spiral spring has a mass of 25 g
  attached to its end. An extra 0.5 g stretches
  the spring 4.9 cm. This extra mass is removed
  and the mass is set vibrating with an amplitude
  of 10 cm. Find
• (c) The maximum velocity

vmax occurs when x 0
vmax ?A
2 x 10-1 ms-1
12
Examples of SHM-Mass of a Spring

• Example A light spiral spring has a mass of 25 g
  attached to its end. An extra 0.5 g stretches
  the spring 4.9 cm. This extra mass is removed
  and the mass is set vibrating with an amplitude
  of 10 cm. Find
• (d) The maximum acceleration

amax -?2xmax
-(2)2 x 10-1
-4 ? 10-1 ms-2
13
Examples of SHM-Simple Pendulum

• Consider a bob of mass m displaced an angle ?
  from the vertical.

Resolving force tangentially
mgsin? ma
a gsin?
If ? is small ( lt15o) then ? sin?
a g?
But ? x/L
a g/L x
Comparing with a -w2x
? The body exhibits SHM
i.e.
14
Energy of a Simple Harmonic Oscillator

• The total mechanical energy of the particle is
  the sum of its potential and kinetic energies
• ie E K U
• K ½ mv2 at any instant
• U work done against restoring force to move
  the mass to position x

Total energy at any position x is given by E
½mv2 ½Kx2
15
Energy of a Simple Harmonic Oscillator

• Now v2 ?2A2 - x2 and K m?2

Then E ½m?2A2 - x2 ½ m?2x2
½m?2A2
½KA2
ie. a constant
ie. in oscillatory motion the energy is
constantly changing from kinetic to potential
energy and back again
16
Energy of a Simple Harmonic Oscillator
at x A energy is purely potential
x 0 energy is purely kinetic
at x A energy is purely potential
at any other point the energy is partially
kinetic and partially potential.
17
Energy of a Simple Harmonic Oscillator

• Exercise A 5.00 kg object connected to a
  massless spring for force constant 20.0 N/m
  oscillates on a horizontal, frictionless track.
  Calculate the total energy of the system and the
  maximum velocity of the object if the amplitude
  of the motion is 3.00 cm.

E K U
When the Object is at x A
E U
9.00?10-3 J
½(20)(3?10-2)2
E ½KA2
When the Object is at x 0
E K ½mv2
v ?(2E/m)
?(2? 9.00?10-3 /5.00) 0.06 m/s
?