Vibrations and Waves
This chapter covers oscillatory motion and wave propagation.
Hooke’s Law
The behavior of many springs approximates Hooke’s Law:
F = -kx
x is the amount by which the spring is stretched or compressed from its equilibrium
position and F is the restoring force exerted by the spring. The negative sign means that
the force is opposite to the direction of the stretch or compression. k is the force constant
of the spring (N/m) and is a measure of the spring stiffness.
Example:
A mass, m = 500 g, is hung from the end of a spring attached to a support. In its
equilibrium position, the mass stretches the spring by 5 cm. What is the force constant of
the spring?
k = F/x = mg/x = (0.5 kg)(9.8 m/s2)/(0.05 m) = 98/m
Elastic Potential Energy
In a previous chapter, it was shown that the potential energy associated with the stretch or
compression of a Hooke’s law spring is
U = ½ kx2
In the above example, the potential energy stored in the spring is U = ½ (98/m)(0.1 m)2 =
0.49 J
Spring oscillation
Assume that one end of a horizontal spring is anchored to a support and a mass is
attached to the other end. The mass is pulled a distance x = A from its equilibrium
position and allowed to oscillate back and forth on a frictionless surface. The position of
the mass will be given as a function of time is given by
x  A cos( 2 ft )
where the frequency of oscillation is given by
f 
1
2
k
m
1
The period T of oscillation is 1/f, so
T  2
m
k
The oscillation of a Hooke’s law spring is called Simple Harmonic Motion (SHM).
A
A
A
The energy of the oscillating mass is given by
E  12 mv 2  12 kx2
At t = 0, v = 0 and x = A, so E  12 kA2  12 mv 2  12 kx2
Solving for v, we have v  
k 2
(A  x2 )
m
So, when x = A, v = 0. And when x = 0, v  
Since F = ma = -kx, then a  
k
x.
m
2
k
A
m
Thus, the acceleration is always opposite to the displacement and its magnitude is a
maximum when the displacement is a maximum.
Assuming that the mass is released from rest at x = A , the expressions for x, v, and a are
given by
x  xmax cos( t )  A cos(2 ft)
v  vmax sin(  t )  vmax sin( 2 ft)
a  amax cos( t )  amax cos( 2 ft)
The maximum v and a are given by
vmax   A
a max   2 A
The frequency (f), angular frequency (), and period (T) are related by
f 
1


T 2
x
A
T/2
t
T
-A
v
vmax
T/2
t
T
-vmax
a
amax
T
t
T/2
-amax
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Example:
A 0.25-kg mass is attached to a spring with force constant k = 36 N/m. The mass is
displaced 5 cm from its equilibrium position and released from rest.
What is the frequency and period of oscillation?
f 
1
2
36 N / m
1
 1.91 Hz , T   0.52 s
0.25kg
f
What is the maximum speed of the mass?
vmax   A  2 fA  2 (1.91Hz )(0.05m)  0.6 m / s (occurs at x = 0)
What is the magnitude of the maximum acceleration?
amax   2 A  (2 f ) 2 A  (2 (1.91Hz )) 2 (0.05m)  7.2 m / s 2
(occurs at x =  5 cm)
How long does it take for the mass to first return to x = 0?
t  14 T  14 (0.52 s)  0.13 s
Pendulum
A simple pendulum consists of a mass suspended
from the end of a light spring. When displaced a
small amount from equilibrium it will undergo simple
harmonic motion, somewhat like a mass at the end of
a Hooke’s law spring.
The restoring force acting on the mass is the
component of mg tangent to arc of the circle in which
it swings.
Fs  mg sin 
.
For small angles, sin   ( in radians). So, using  = s /L,
Fs  mg   mg
s
 mg 
 
s
L
 L 
This is like Hooke’s law, where k = mg/L. Thus,
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
T
L
s
mg sin
mg cos
mg
T  2
m
m
 2
k
mg / L
T  2
L
g
Example:
A 100-g mass oscillates at the end of a 50-cm string. What is its period?
0 .5 m
T  2
9 .8 m / s 2
 1.42 s
What would be the length of a pendulum that had a period of 1 s?
L
gT 2
4
2

(9.8)(1)
4 2
 0.25 m
How would the period change if the mass were doubled?
No change since T is independent of mass.
How would the period change is a pendulum were taken to the moon?
T would increase since g(moon) < g(earth).
How would the period change if the amplitude (maximum swing) were increased?
No change as long as  is sufficiently small such that   sin.
Waves
A traveling wave is a disturbance that propagates in space and time. It can be a single
pulse or a periodic disturbance.
For a transverse wave the disturbance is perpendicular to the direction of motion. For
longitudinal wave, the disturbance is along the direction of motion.
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As an example, a longitudinal wave could be established in a stretched slinky by pushing
and pulling one end in an oscillatory manor. The regions of compression and expansion
would propagate along the length of the slinky.
v
A transverse wave is shown below. Such a wave could be established on a rope by
moving one end up and down.
v
The wave speed is given by
v
F

where F is the tension in the rope in N and  is the mass per unit length of the rope in
kg/m.
Example:
A string clamped at one is held in tension by a 1.5-kg
weight attached to the other, as shown to the right.
The mass of the string is 50 g and the length is 2 m.
What is the speed of a wave on this string?
F  Mg  (1.5kg)(9.8m / s 2 )  14.7 N
  m / L  0.05kg / 2m  0.025 kg / m
v
F


14.7 N
 24.2 m / s
0.025kg / m
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Wave equation:
A “sinusoidal” wave traveling to the right can be described by the equation
y  A sin( kx   t   )
where
A = amplitude = maximum vertical displacement (m)
k = 2/ = wave number (m-1)
 = wavelength (m)
 = 2f = 2/T = angular frequency (rad/s)
f = frequency (Hz)
T = period (sec)
 = phase angle (Depends on choice of t = 0.)
The wave speed can be written as
v
x 
  f
t T
y
A

x
Example:
A wave is described by the equation y  0.05m sin( 2 x  5 ) . What are the wavelength,
period, and wave speed?
k = 2/ = 2,  =  = 3.14 m
 = 2/T = 5, T = 0.4 = 1.26 s
v = /T = 3.14m/1.26s = 2.5 m/s
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Interference and Reflection of Waves
Waves obey the principle of superposition. This means that when two waves meet, the
resultant wave is obtained by adding the two waves point by point. Waves do not collide
and bounce off each other as is the case for particles. They travel through each other and
emerge as though they did not interact.
Waves can reflect, however, from the end of a rope. If the end of the rope is tightly
secured, then the wave will invert when reflected. If the end of the rope is untied and
loose, then the wave will reflect back without inversion. If two ropes with different mass
densities are tied together, then a wave will be partly reflected from the point where they
are joined together.
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