Homework: chapter 16
Wave Motion
1,2,3,4,10,13,14,17,19,21,24,27,30,32,34,36,
40, 43,46,52,59,62
1. At t = 0, a transverse pulse in a wire is described by the function
6
y 2
x 3
where x and y are in meters. Write the function y(x, t) that describes this
pulse if it is traveling in the positive x direction with a speed of 4.50 m/s.
Replace x by
to get
x  vt x  4.5t
y
6
( x  4.5t ) 2  3
2. Ocean waves with a crest-to-crest distance of 10.0 m can be described
by the wave function
y(x, t) = (0.800 m) sin[0.628(x – vt)]
where v = 1.20 m/s.
(a) Sketch y(x, t) at t = 0.
(b) Sketch y(x, t) at t = 2.00 s. Note how the entire wave form has
shifted 2.40 m in the positive x direction in this time interval.
3. A pulse moving along the x axis is described by
 x  5.00t 
y x , t   5.00e
2
where x is in meters and t is in seconds.
Determine:
(a) the direction of the wave motion,
(b) (b) the speed of the pulse.
The given function is of the form f(X+vt)
It describes a wave moving to the left at
.
v  5.00 m s
4. Two points A and B on the surface of the Earth are at the same longitude and
60.0 apart in latitude. Suppose that an earthquake at point A creates a P
wave that reaches point B by traveling straight through the body of the Earth
at a constant speed of 7.80 km/s. The earthquake also radiates a Rayleigh
wave, which travels along the surface of the Earth in an analogous way to a
surface wave on water, at 4.50 km/s.
(a) Which of these two seismic waves arrives at B first?
(b) What is the time difference between the arrivals of the two waves at B? Take
the radius of the Earth to be 6 370 km.
a) The longitudinal wave is traveling a shorter distance and is faster, so it will
arrive at B first
(b)
The wave that travels through the Earth must travel
a distance of


2R sin 30.0  2 6.37  106 m sin 30.0  6.37  106 m
At a speed of 7800 m/s
Then, it takes
6.
37  106 m
 817 s
7 800 m s
The wave that travels along the Earth’s surface must travel


a distance of
s R  R 
rad  6.67  106 m
3

16.4: continuation
at a speed of
4 500 m/s
Therefore, it takes
Time difference:
6.
67  106
 1 482 s
4 500
665 s  11.1 m in
10. A sinusoidal wave on a string is described by
y = (0.51 cm) sin(kx – t)
where k = 3.10 rad/cm and  = 9.30 rad/s. How far does a wave crest move in
10.0 s? Does it move in the positive or negative x direction?
y   0.0051 m  sin  310x  9.30t
v
s vt

k

9.
30
 0.
030 0 m s
310
0.
300 m in positive x-direction
13. A sinusoidal wave train is described by
y = (0.25 m) sin(0.30x – 40t)
where x and y are in meters and t is in seconds. Determine for this wave the
(a) amplitude,
(b) angular frequency,
(c) angular wave number,
(d) wavelength,
(e) wave speed, and
(f) direction of motion.
(a
A  0.250 m
(b)
  40.0 rad s
(c )
k
(d)

(e)
 40.0 rad s

v  f      
  20.9 m   133 m s
 2 

2
0.
300 rad m
2
2


k
0.
300 rad. m
(f) The wave moves to the right,
20.
9m
in  x direction
14.
(a) Plot y versus t at x = 0 for a sinusoidal wave of the form
y = (15.0 cm) cos(0.157x – 50.3t) , where x and y are in centimeters and t is
in seconds.
(b) Determine the period of vibration from this plot and compare your result with
the value found in Example 16.2.
T 
2


2

50.
3
0.
125 s
This agrees with the period found in the example in the text.
17. A transverse wave on a string is described by the wave function
y  (0.120m) sin[( x / 8  4 t )]
(a) Determine the transverse speed and acceleration at t = 0.200 s for the
point on the string located at x = 1.60 m.
(b) What are the wavelength, period, and speed of propagation of this wave?
::
(a)
v
a
k

8
y

 (0.12)( 4 ) cos( x:  4 t )
t
8
dv
dt

2

a   0.120 m
 4  2 sin 


x  4 t

8
  16.0 m
2
T

16.
0m
v


T
0.
500 s
  4 
T  0.500 s
32.
0 m s
19. A sinusoidal wave of wavelength 2.00 m and amplitude 0.100 m travels on
a string with a speed of 1.00 m/s to the right. Initially, the left end of the string
is at the origin. Find: (a) the frequency and angular frequency,
(b) the angular wave number, and
(c) the wave function for this wave. etermine the equation of motion for
(d) the left end of the string, and
(e) the point on the string at x = 1.50 m to the right of the left end. (f) What is
the maximum speed of any point on the string?
(a)
v 1.00 m s
f 
 0.500 H z

2.00 m
  2 f  2  0.500 s  3.14 rad s
2
2
 3.14 rad m
2.00 m
(b)
k
(c )
y  A si
n  kx  t  
y
e)


 0.100 m  sin  3.14x m
y   0.
100 m
 3.14t s 0
 sin  4.71 rad  3.14t s
y
vy   0.100 m   3.14 s cos 3.14x m  3.14t s
t

The cosine varies between +1
314 m
and –1, so vy  0.
s
21. A telephone cord is 4.00 m long. The cord has a mass of 0.200 kg. A
transverse pulse is produced by plucking one end of the taut cord. The pulse
makes four trips down and back along the cord in 0.800 s. What is the
tension in the cord?
The down and back distance is
4.00 m  4.00 m  8.00 m
The speed is then
v
Now,
so
4 8.00 m 
dtot
. al

 40.0 m s 
t
0.800 s

T

0.200 kg
 5.00  102 kg m
4.00 m

T  v2  5.00  102 kg m

 40.0 m s  80.0 N
2
24. A transverse traveling wave on a taut wire has an amplitude of 0.200 mm and
a frequency of 500 Hz. It travels with a speed of 196 m/s.
(a) Write an equation in SI units of the form y = A sin(kx – t) for this wave.
(b) The mass per unit length of this wire is 4.10 g/m.
Find the tension in the wire.
  2 f  2  500  3140 rad s

3 140
k

 16.
0 rad m
v
196


y  2.00  104 m sin 16.0x  3140t
b)
v  196 m
T  158 N
s
T
4.
10  103 kg m
27. Transverse waves travel with a speed of 20.0 m/s in a string under a
tension of 6.00 N. What tension is required for a wave speed of 30.0 m/s in the
same string?
Since µ is constant,

T2
2
v2

T1
v12
2
and
 30.0 m s
 v2 
T2    T1  

 v1 
 20.0 m s
.
2
 6.00 N  
13.5 N
30. Review problem. A light string with a mass per unit length of 8.00 g/m has
its ends tied to two walls separated by a distance equal to three-fourths the
length of the string (Fig. P16.30). An object of mass
m is suspended from the center of the string, putting a tension in the string.
(a) Find an expression for the transverse wave speed in the string as a function of
the mass of the hanging object.
(b) What should be the mass of the object suspended from the string in order to
produce a wave speed of 60.0 m/s?
m g  2T si
n
mg
2si
n
T 
cos 
3L
8
L
2

3
4
  41.
4
a)

mg

2 sin 41.4 

v
v
T




kg m sin 41.4 
9.80 m s2
2 8.00  103

m s
v   30.4

kg 

v  60.0  30.4 m
and

m
m
m  3.89 kg
Figure P16.30
32. Review problem. A light string of mass m and length L has its ends tied to
two walls that are separated by the distance D. Two objects, each of mass M,
are suspended from the string as in Figure P16.32. If a wave pulse is sent from
point A, how long does it take to travel to point B?
Refer to the diagrams. From the free-body
diagram of point A:
 Fy  0  T1 sin  M g
 Fx
 0  T1 cos  T
Combining these equations to eliminate T1 gives the tension
in the string connecting points A and B:
Figure P16.32
M g
T 
t
an 
The speed of transverse waves in this segment
of string is then
Time for the pulse to travel
v
T


Mg
tan
m
L

M gL
m tan 
from A to B
t
L
2
v

m L tan 
4M g
34.
A taut rope has a mass of 0.180 kg and a length of 3.60 m. What
power must be supplied to the rope in order to generate sinusoidal waves
having an amplitude of 0.100 m and a wavelength of 0.500 m and traveling with
a speed of 30.0 m/s?
f
v


30.0
 60.0 H z
0.500
  2 f  120 rad s
power
P
1
1  0.180
2
2
 2A 2v  
120   0.100  30.0  1.07 kW


2
2  3.60 
36.
Transverse waves are being generated on a rope under constant
tension. By what factor is the required power increased or decreased if (a) the
length of the rope is doubled and the angular frequency remains constant, (b)
the amplitude is doubled and the angular frequency is halved, (c) both the
wavelength and the amplitude are doubled, and (d) both the length of the rope
and the wavelength are halved?
T is constant
v
T

P 
P i
s const
ant
(a) If L is doubled, v remains constant and
.
.
(b) If A is doubled and  is halved,
(c) If  and A are doubled, the product
(d)
If L and  are halved, then
1
 2A 2v
2
P   2A 2 rem ains constant
 A 
2
P
2
A2

2
remains constant,
So p remains constant
i
s quadrupl
ed
(Changing L doesn’t affect p)
40. The wave function for a wave on a taut string is
y(x, t) = (0.350 m)sin(10t – 3x + /4)
where x is in meters and t in seconds.
(a) What is the average rate at which energy is transmitted along the string if the
linear mass density is 75.0 g/m?
(b) What is the energy contained in each cycle of the wave?


y  0.
35si
n  10 t 3 x 


4
Comparing
with
k
Then:
(a)
(b)
y  A sin  kx  t    A sin t kx     
3
m
  10 s A  0.35 m
10 s




 3.
33 m s
2
k
3 m
The rate of energy transport is
1
1
2
2
P   2A 2v  75 103 kg m 10 s  0.35 m  3.33 m s  15.1 W
2
2
v  f  2 f


The energy per cycle is
1
1
2
2 2 m
E  P T   2A 2  75  103 kg m 10 s  0.35 m 
 3.02 J
2
2
3


43.
(a) Evaluate A in the scalar equality
(7+3)4=A
(b) Evaluate A, B, and C in the vector equality .





Explain
7.0ihow
 3you
.0karrive
 Aiatthe
Bjanswers
 Ck to convince a student who thinks that
you cannot solve a single equation for three different unknowns.
(c) What If? The functional equality or identity
A + B cos (Cx + Dt + E) = (7.00 mm) cos(3x + 4t + 2) is true for all values of
the variables x and t, which are measured in meters and in seconds,
respectively. Evaluate the constants A, B, C, D, and E. Explain how you
arrive at the answers.
A   7.
00  3.
00 4.
00

A  40.0
b) A=7.0, B=0 and C=3.0
(c)
In order for two functions to be identically equal, they must be equal
for every value of every variable. They must have the same graphs. In
A  B cos Cx  D t E  0  7.00 m m cos 3.00x  4.00t 2.00
16.43 continuation
the equality of average values requires that
The equality of maximum values requires
A 0
B  7.00 m m
The equality for the wavelength or periodicity as a function of x requires . The
equality of period requires C  3.00 rad m
The equality of period requires
D  4.00 rad s
And the equality of zero-crossings requires
E  2.00 rad
46. (a) Show that the function y(x,t) = x2 + v2 t2 is a solution to the wave
equation.
(b) Show that the function in part (a) can be written
as f(x + vt) + g(x – vt), and determine the functional forms for f and g.
(c) What If? Repeat parts (a) and (b) for the function y(x,t) = sin(x)cos(vt).
y  x2  v2t2
y
 2x
x
 2y
2
y
 v2 2t
t
2
 y
t2
x
 2y
2
t2
2
 2v2
1  y
 2
v t2
1
1
2
x

vt



 x  vt 2
2
2

follows
1 2
1
1
1
x  xvt v2t2  x2  xvt v2t2
2
2
2
2
 x2  v2t2
so
f x  vt 
1
 x  vt 2
2
g x  vt 
As required
1
 x  vt 2
2
y  si
n xcosvt
Part c) of 16.47
y
 cosx cosvt
x
 2y
x2
y
  vsin x sin vt
t
Then:
 2y
Note
x2
 2y
2
t
1  2y
 2
v t2
  sin x cosvt
 v2 sin x cosvt
becomes
So
 sin xcosvt
1 2
v sin xcosvt
2
v
which is true as required.
.
sin  x  vt  sin x cosvt cosx sin vt
sin  x  vt  sin x cosvt cosx sin vt
sin x cosvt f x  vt  g x  vt
f x  vt 
1
sin  x  vt
2
and
g x  vt 
1
sin  x  vt
2
52.
Review problem. A block of mass M, supported by a string, rests
on an incline making an angle 
with the horizontal (Fig. P16.52). The length of the string is L and its mass is m
<<M. Derive an expression for the time interval required for a transverse wave
to travel from one end of the string to the other.
Assuming the incline to be frictionless
and taking the positive x-direction to
be up the incline:
 Fx  T  M gsin  0
or the tension in the string is
T  M gsin
The speed of transverse waves in the string is then
M gL sin 
m

m
L
The time interval for a pulse to travel the string’s length is
v
T

M g sin 
Fig. 16.52
t
L
L
v

m

M gL sin 
mL
M g sin 
59. A rope of total mass m and length L is suspended vertically. Show that a
transverse pulse travels the length of the rope in a time interval . (Suggestion: First
find an expression for the wave speed at any point a distance x from the lower end
by considering the tension in the rope as resulting from the weight of the segment
below that point.)
v
T
T  xg

Then
But
dx
v
dt
L
t

0
the weight of a length x, of rope.
v
dt
dx

gx
1
g
gx
dx
gx
x
1
2
L

0
2
L
g
62.
Determine the speed and direction of propagation of each of the
following sinusoidal waves, assuming that x is measured in meters and t in
seconds.
(a)
y = 0.60 cos(3.0x – 15t + 2)
(b)
y = 0.40 cos(3.0x + 15t – 2)
(c)
y = 1.2 sin(15t + 2.0x)
(d)
y = 0.20 sin[(12t – (x/2) + ]
a)
b)
c)
d)
v

k
v

15.0
 5.00 m s in positive x-direction
3.00
15.
0

3.
00
5.
00 m s i
n negati
ve x-di
recti
on
15.0
v
 7.50 m s in negative x-direction
2.00
v
12.
0
1
2

24.
0 m s in positive x-direction