QUICK QUIZ 16.1
(end of section 16.2)
While sitting at the beach,
you count the number of
waves that hit the beach
during a certain amount of
time. This measurement
is most closely associated
with a) the period of the
waves, b) the frequency of
the waves, c) the
wavelength of the waves,
or d) the speed of the
waves.
QUICK QUIZ 16.1 ANSWER
(b). The frequency of a wave is associated with the number of
cycles of the wave that occur in a given amount of time. When
we count the waves that hit the beach, we are actually counting
the number of wave crests.
QUICK QUIZ 16.2
(end of section 16.2)
In Equation 16.10, y = A sin(kx – wt), for a traveling
sinusoidal wave, the variable, k, is related to the
spring constant, k, from Chapter 15 a) in the sense
that it is related to a force, b) in the sense that it is
associated with a displacement, c) in the sense that it
is associated with oscillatory motion, or d) is not
related to the spring constant, k.
QUICK QUIZ 16.2 ANSWER
(d). Unfortunately, when representing the large quantity of
physical variables that exist, the alphabet becomes quickly
exhausted. The variable k is used for two completely different
quantities, the spring constant and the wave number. These
quantities must be completely different since the units are
completely different.
QUICK QUIZ 16.3
(end of section 16.3)
You suspend an object from the end of a hanging rubber band, send a pulse along the
band and measure the speed of the pulse to be v. You then quadruple the mass of the
object that you hang on the rubber band and the rubber band’s length increases by a
factor of two from its original length with the first object. If you now send a pulse
along the band, the speed of the pulse will be a) v/(22) b) v/2, c) v/2, d) v, e) 2 v,
f) 2v, or g) (22)v
QUICK QUIZ 16.3 ANSWER
(g). By increasing the mass of the hanging object by
a factor of four, you have increased the tension by a
factor of four. Since the rubber band has doubled its
length, its mass per unit length or linear mass density
has gone down by a factor of two. Therefore,
v1 
T

and
v2 
4T
1
2


8T

2 2
T

 (2 2 )v1.
QUICK QUIZ 16.4
(end of section 16.5)
You perform an experiment on a string and generate
sinusoidal waves of an amplitude, A, and frequency, f.
You then perform a similar experiment on a string that
has twice the linear mass density and which is under half
the tension as the original string. To generate sinusoidal
waves of an amplitude, A, and frequency, f, in this new
string, you will have to transfer energy to the new string
at a rate that is a) one fourth the rate for the original
string, b) half the rate for the original string, c) the same
as the rate for the original string, d) twice the rate for the
original string, or e) four times the rate for the original
string.
QUICK QUIZ 16.4 ANSWER
(c). Equation 16.21 relates the energy rate transfer to the
linear mass density, frequency, amplitude and speed. The
speed is in turn related to the tension and the linear mass
density so that the equation may be rewritten,
P
1
2
w A v 
2
2
1
2
 4 f A
2
2
2
T

 2 2f 2 A2
 T.
Therefore, if one doubles the linear mass density and
halves the tension, the rate of energy transfer must remain
the same to keep the same amplitude and frequency.
QUICK QUIZ 16.5
(end of section 16.6)
Which of the following is a solution to the linear
wave equation? a) y = x + vt2, b) y = sin2(x + vt), c)
both a and b, d) neither a nor b.
QUICK QUIZ 16.5 ANSWER
(b). Only functions of the form y = f(x ± vt) are solutions to the wave equation.
The function y = x + vt2 is not of this form and it is easy to verify that it is not a
solution. We have,
2y 2
1 2y 1 2y
2v 2
2
2

(x

vt
)

0
,
and

(x

vt
)

 .
2
2
2
2
2
2
2
x
x
v t
v t
v
v
Since 0 is not, in general, equal to 2/v, the function is not a solution to the wave
equation. On the other hand, y = sin2(x + vt) is of the form y = f (x ± vt) and is
a solution to the wave equation. We have,
2y 2

2

sin
(x

vt)

(2sin(x  vt)cos(x  vt))  2(cos 2 (x  vt)  sin 2 (x  vt) )
2
2
x
x
x
1 2y 1 2
1 
and 2 2  2 2 sin 2 (x  vt)  2 2vsin(x  vt)cos(x  vt) 
v t
v t
v t
1
2
2
2
2
2v(vcos
(x

vt)
v
sin
(x

vt))

2
(cos
(x

vt)

sin
(x  vt) ).
2
v