Chapter 13: Vibrations and Waves
Suggested homework problems:12,33,47,54,58
Hooke’s Law
 Hooke’s
law and oscillation
• A simple example of vibration motion: an object attached to a spring.
Fs  kx
The negative sign means that the force
exerted by the spring is always directed
opposite the displacement of the object
(restoring force).
A restoring force always pushes or pulls
the object toward the equilibrium position.
No friction
Hooke’s Law
 Hooke’s
law and oscillation (cont’d)
• A simple example of vibration motion:
an object attached to a spring.
Suppose the object is initially pulled
a distance A to the right and released
from rest. Then the object does simple
harmonic motion.
No friction
Simple harmonic motion occurs when the
net force along the direction of motion
obeys Hooke’s law – when the net force
is proportional to the displacement from
equilibrium point and is always
directed toward the equilibrium
point.
periodic motion
Hooke’s Law
 Hooke’s
law and oscillation (cont’d)
• A simple example of vibration motion:
an object attached to a spring.
No friction
Terminology:
- The amplitude A is the maximum
distance of the object from its equilibrium position. –A<= x <= A.
- The period T is the time it takes the
object to move through one complete
cycle of motion from x=A to x=-A and
then back to x=A.
- The frequency f is the number
of complete cycles or vibrations
per unit time. f=1/T.
periodic motion
Hooke’s Law
 Hooke’s
law and oscillation (cont’d)
• Example 13.1 : Measuring the spring constant
F  F
g
 Fs  mg  kd  0
mg
k
d
• Harmonic oscillator equation
ma  F  kx
k
a x
m
Elastic Potential Energy
 Elastic
potential energy
• The energy stored in a stretched or compressed spring
1 2
PEs  kx
2
• The conservation of energy:
If there are only conservative forces
( KE  PEg  PEs ) i  ( KE  PEg  PEs ) f
If there are also non-conservative forces
Wnc 
( KE  PEg  PEs ) f  ( KE  PEg  PEs )i
Elastic Potential Energy
 Elastic
potential energy
• Conservation of energy :
1 2 1 2 1 2
kA  mv  kx
2
2
2

k 2
v
A  x2
m

- The velocity is zero at x=+A,-A.
- The velocity is at its maximum
at x=0
Comparing Simple Harmonic Motion with
Uniform Circular Motion
 Uniform
circular motion
• A circular motion and its projection
As the turntable rotates with constant
angular speed, the shadow of the ball
moves back and forth with simple
harmonic motion.
v
sin   
v0
A2  x 2
A
v0 A2  x 2
v
A
c.f. simple harmonic
motion
v

k 2
A  x2
m

Comparing Simple Harmonic Motion with
Uniform Circular Motion
 Period
and Frequency
• Period of oscillation (T)
One period is completed when the ball rotates 360o and
moves a distance v0T.
v0T  2A
2A
T
v0
From conservation of energy, for simple harmonic oscillation of
a spring system with the spring constant k at x=0,
1 2 1 2
A
m
kA  mv0  
2
2
v0
k
T  2
k
m
Comparing Simple Harmonic Motion with
Uniform Circular Motion
 Period
and Frequency
• Frequency (f) and angular frequency (w)
Frequency is how many complete rotations/cycles a simple harmonic
oscillation or uniform circular motion makes per unit time.
f 
1
1

T 2
k
m
Units : hertz (Hz) = cycles per second
Angular frequency is a frequency measured in terms of angle.
w  2f 
k
m
Position, Velocity, and Acceleration
as a Function of Time
x
vs. time
• We can obtain an expression for the position of an object with
simple harmonic motion as a function of time.
x  A cos
  wt
if constant angular speed
x  A cos(wt )
w
 2

 2f
t
T
x  A cos( 2ft )
Position, Velocity, and Acceleration
as a Function of Time
v
vs. t
• We can obtain an expression for the velocity of an object with
simple harmonic motion as a function of time.
v0
v
A
A2  x 2
x  A cos(wt ) if constant angular
speed
v0
w 1
f 
 
2 T 2A
v0
v
A2  A2 cos 2 (wt )  v0 sin( wt )
A
  Aw sin( wt )   Aw sin( 2ft )
v   Aw sin( 2ft )
Position, Velocity, and Acceleration
as a Function of Time
 Period
and Frequency
• We can obtain an expression for the position of an object with
simple harmonic motion as a function of time.
sinusoidal
k
a   x  wx
m
x  A cos(wt ) if constant angular
speed
a   Aw 2 cos(2ft)
Motion of a Pendulum
 Pendulum
• If a force is a restoring one, from an analogy of a Hooke’s law
we can prove that the system under influence the force makes
simple harmonic oscillation.
s
s
 mg 
Ft  mg sin   mg sin    mg  
 s  sin    if   1
L
L
 L 
In an analogy to Hooke’s law Ft=-kx,
mg
k
L
k
w  2f 

m
g
L
L
T  
g
The motion of a pendulum is not simple
harmonic in general but it is if the angle
 is small.
Ft=
Motion of a Pendulum
 Physical
pendulum
• In general case, the argument for a pendulum system of a mass
attached to a string can be used to an object of any shape.
I
T  2
mgL
I: moment of
inertia
For a simple pendulum,
I  mL2
mL2
L
T  2
 2
mgL
g
Damped Oscillation
 Oscillation
with friction
• In any real systems, forces of frictions retard the motion induced
by restoring forces and the system do not oscillate indefinitely.
The friction reduces the mechanical energy of the system as time
passes, and the motion is said to be damped.
Waves
 Examples
and sources of waves
• The world is full of waves: sound waves, waves on a string,
seismic waves, and electromagnetic waves such as light, radio
waves, TV signals, x-rays, and g-rays.
• Waves are produced by some sort of vibration:
Vibration of vocal cords, guitar strings, etc sound
Vibration of electrons in an antenna, etc
Vibration of water
 Types
radio waves
water waves
of waves
• Transverse waves
 The bump (pulse) travels to the right
with a definite speed: traveling wave
 Each segment of the rope that is disturbed
moves in a direction perpendicular to the
wave motion: transverse wave
Waves
 Types
of waves (cont’d)
• Longitudinal waves
 The elements of the medium undergo displacements parallel
to the direction of wave motion: longitudial wave
 Their disturbance corresponds to a series of high- and lowpressure regions that may travel through air or through any
material medium with a certain speed.
sound wave = longitudinal wave
C = compression
R = rarefaction
Waves
 Types
of waves (cont’d)
• Longitudinal-transverse waves
Frequency, Amplitude, and
Wavelength
 Frequency,
amplitude, and wavelength
• Consider a string with one end connected to a blade vibrating
according to simple harmonic oscillation.
Amplitude A:
The maximum distance the
string moves.
Wavelength l:
The distance between two
successive crests
Wave speed v:
v=x/t=l/T
wavelength/period)
Frequency f:
v=l/T=fl
wavelength/period)
Frequency, Amplitude, and
Wavelength
 Examples
• Example 13.8: A traveling wave
A wave traveling in the positive x-direction. Find the amplitude,
wavelength, speed, and period of the wave if it has a frequency
of 8.00 Hz. x=40.0 cm and y=15.0 cm.
A  y  15.0 cm  0.150 m
l  x  40.0 cm  0.400 m
v  fl  (8.00 Hz)(0.400 m)
 3.20 m/s
1
1
T 
s  0.125 s
f 8.00
Frequency, Amplitude, and
Wavelength
 Examples
(cont’d)
• Example 13.9: Sound and light
A wave has a wavelength of 3.00 m. Calculate the frequency of the
wave if it is (a) a sound wave, and (b) a light wave. Take the speed
of sound as 343 m/s and that of light as 3.00x108 m/s.
v 343 m/s
(a)
f 
(b)
l

3.00 m
 114 Hz
3.00 108 m/s
f  
 1.00 108 Hz
l
3.00 m
c
Speed of Waves on Strings
 Speed
of waves on strings
• Two types of speed:
 The speed of the physical string that vibrates up and down
transverse to the string in the y-direction
 The rate at which the disturbance propagates along the length
of the string in the x-direction: wave speed
• For a fixed wavelength, a string under greater tension F has a
greater wave speed because the period of vibration is shorter,
and the wave advances one wavelength during one period.
A string with greater mass per unit length m (linear density) vibrates
more slowly, leading to a longer period and a slower wave speed.
v
F
m
Dimension analysis:
[F]=ML/T2, [m]=M/L, F/m=L2/T2, [F/m]=L/T=[v]
Speed of Waves on Strings
 Example
13.10
• A uniform string has a mass M of 0.0300 kg and a length L of 6.00 m.
Tension is maintained in the string by suspending a block of mass
m = 2.00 kg from one end.
(a) Find the speed of the wave.
F  F  mg  0  F  mg
v
F
m

mg
 62.6 m/s
M /L
(b) Find the time it takes the
pulse to travel from the wall
to the pulley.
d 5.00 cm
t 
 0.0799 s
v 62.6 m/s
Interference of Waves
 Superposition
principle
• Tow traveling waves can meet and pass through each other without
being destroyed or even altered.
• When two or more raveling waves encounter each other while
moving through a medium, the resultant wave is found by adding
together the displacements of the individual waves point by point.
 Interference
constructive interference
(in phase)
destructive interference
(out of phase)
Interference of Waves
 Example
13.10
• A uniform string has a mass M of 0.0300 kg and a length L of 6.00 m.
Tension is maintained in the string by suspending a block of mass
m = 2.00 kg from one end.
(a) Find the speed of the wave.
F  F  mg  0  F  mg
v
F
m

mg
 62.6 m/s
M /L
(b) Find the time it takes the
pulse to travel from the wall
to the pulley.
d 5.00 cm
t 
 0.0799 s
v 62.6 m/s
Reflection of Waves
 Reflection
of waves at a fixed end
Reflected wave is inverted
Reflection of Waves
 Reflection
of waves at a free end
Reflected wave is not inverted