Chapter 10: Oscillations
•Simple Harmonic Motion
•The Pendulum
•Damped Oscillations, Forced Oscillations, and Resonance
§10.5 Simple Harmonic Motion
Simple harmonic motion (SHM)
occurs when the restoring force
(the force directed toward a stable
equilibrium point) is proportional to
the displacement from equilibrium.
The motion of a mass on a spring is an example of SHM.
Equilibrium
position
y
x
x
The restoring force is F=-kx.
Assuming the table is frictionless:
F
x
  kx  max
k
a x t    xt 
m
Also,
1
1
2
2
E t   K t   U t   mvt   kxt 
2
2
At the equilibrium point x=0 so a=0 too.
When the stretch is a maximum, a will be a maximum too.
The velocity at the end points will be zero, and it is a
maximum at the equilibrium point.
§10.6-7 Representing Simple Harmonic
Motion
When a mass-spring system is oriented vertically, it will
exhibit SHM with the same period and frequency as a
horizontally placed system.
SHM
graphically
A simple harmonic oscillator can be described mathematically
by:
xt   A cos t
x
vt  
  A sin t
t
v
at  
  A 2 cos t
t
Or by:
xt   A sin t
x
vt  
 A cos t
t
v
at  
  A 2 sin t
t
where A is the amplitude
of the motion, the
maximum displacement
from equilibrium, A=vmax,
and A2 =amax.
The period of oscillation is T 
2

.
where  is the angular frequency of the
oscillations, k is the spring constant and
m is the mass of the block.

k
m
Example (text problem 10.28): The period of oscillation of an
object in an ideal mass-spring system is 0.50 sec and the
amplitude is 5.0 cm. What is the speed at the equilibrium
point?
At equilibrium x=0:
1 2 1 2 1 2
E  K  U  mv  kx  mv
2
2
2
Since E=constant, at equilibrium (x = 0) the
KE must be a maximum. Here v = vmax = A.
Example continued:
The amplitude A is given, but  is not.
2
2


 12.6 rads/sec
T
0.50 s
and v  Aω  5.0 cm 12.6 rads/sec   62.8 cm/sec
Example (text problem 10.41): The diaphragm of a speaker
has a mass of 50.0 g and responds to a signal of 2.0 kHz by
moving back and forth with an amplitude of 1.810-4 m at that
frequency.
(a) What is the maximum force acting on the diaphragm?


2
2


F

F

ma

m
A


mA
2

f

4

mAf

max
max
2
The value is Fmax=1400 N.
2
Example continued:
(b) What is the mechanical energy of the diaphragm?
Since mechanical energy is conserved, E = KEmax = Umax.
1 2
U max  kA
2
1 2
KEmax  mvmax
2
The value of k is unknown so use KEmax.
KEmax
1 2
1
1
2
2
 mvmax  m A   mA2 2f 
2
2
2
The value is KEmax= 0.13 J.
Example (text problem 10.47): The displacement of an object
in SHM is given by:
yt   8.00 cmsin 1.57 rads/sec t 
What is the frequency of the oscillations?
Comparing to y(t)= A sint gives A = 8.00 cm and
 = 1.57 rads/sec. The frequency is:
 1.57 rads/sec
f 

 0.250 Hz
2
2
Example continued:
Other quantities can also be determined:
The period of the motion is
2
2
T

 4.00 sec
 1.57 rads/sec
xmax  A  8.00 cm
vmax  A  8.00 cm 1.57 rads/sec   12.6 cm/sec
amax  A 2  8.00 cm 1.57 rads/sec   19.7 cm/sec 2
2
§10.8 The Pendulum
A simple pendulum is constructed by attaching a mass to
a thin rod or a light string. We will also assume that the
amplitude of the oscillations is small.
A simple pendulum:

An FBD for the
pendulum bob:
L
y
T
m
Assume <<1 radian

w
x
F
x
Apply Newton’s 2nd
Law to the pendulum
bob.
 mg sin   mat
v2
 Fy  T  mg cos   m r
If we assume that <<1 rad, then sin   and cos 1 then
the angular frequency of oscillations is found to be:

g
L
L
The period of oscillations is T  2
g
Example (text problem 10.60): A clock has a pendulum that
performs one full swing every 1.0 sec. The object at the end
of the string weighs 10.0 N. What is the length of the
pendulum?
L
T  2
g


gT 2
9.8 m/s 2 1.0 s 

 0.25 m
Solving for L: L 
2
2
4
4
2
Example (text problem 10.84): The gravitational potential
energy of a pendulum is U=mgy. Taking y=0 at the lowest
point of the swing, show that y=L(1-cos).

Lcos
L
L
y  L(1  cos  )
y=0
A physical pendulum is any rigid object that is free to
oscillate about some fixed axis. The period of oscillation of
a physical pendulum is not necessarily the same as that of
a simple pendulum.
§10.9 Damped Oscillations
When dissipative forces such as friction are not negligible,
the amplitude of oscillations will decrease with time. The
oscillations are damped.
Graphical representations of damped oscillations:
§10.10 Forced Oscillations and
Resonance
A force can be applied periodically to a damped oscillator (a
forced oscillation).
When the force is applied at the natural frequency of the
system, the amplitude of the oscillations will be a maximum.
This condition is called resonance.
Summary
•Stress and Strain
•Hooke’s Law
•Simple Harmonic Motion
•SHM Examples: Mass-Spring System, Simple Pendulum
and Physical Pendulum
•Energy Conservation Applied to SHM