Physics 295 Solutions to Problems in Chapter 16 Wave Motion
P16.4
The distance the waves have traveled is
d  7.80 km s t   4.50 km s  t  17.3 s
where t is the travel time for the faster
wave.
Then,
7.80  4.50  km s t   4.50 km s 17.3 s
or t 
 7.80  4.50 km s
and the distance
is d  7.80 km s  23.6 s  184 km .
P16.5
(a)
Let u  10 t  3 x 
 4.50 km s 17.3 s

4
 23.6 s
du
dx
 10  3
 0 at a point of constant phase
dt
dt
dx 10

 3.33 m s
dt
3
The velocity is in the positive x-direction .
(b)


y  0.100, 0   0.350 m  sin  0.300    0.054 8 m  5.48 cm
4

(c)
k
(d)
vy 
2

 3 :   0.667 m
  2 f  10 : f  5.00 Hz
y


  0.35010  cos 10 t  3 x  
t
4

vy , max  10  0.350  11.0 m s
*P16.11
(a)
  2 f  2  5 s1   31.4 rad s
(b)

k
(c)
v 20 m s

 4.00 m
f
5 s1
2


2
 1.57 rad m
4m
In y  A sin  kx   t    we take A  12 cm . At x  0 and t  0 we
have y  12 cm  sin  . To make this fit y  0 , we take   0 . Then
y   12.0 cm  sin   1.57 rad m  x   31.4 rad s t 
(d)
(e)
The transverse velocity is
y
  A cos kx  t 
t
Its maximum magnitude is
A  12 cm  31.4 rad s  3.77 m s
ay 
vy
t


 A cos kx  t   A 2 sin  kx  t 
t
A 2   0.12 m   31.4 s1   118 m s2
2
The maximum value is
P16.21 The down and back distance is 4.00 m  4.00 m  8.00 m .
Now,  
So
dtotal 4  8.00 m 
T

 40.0 m s 
t
0.800 s

v
The speed is then
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
L
g
P16.26 The period of the pendulum is T  2
Let F represent the tension in the string (to avoid confusion with the period) when
the pendulum is vertical and stationary. The speed of waves in the string is then:
v
F


Mg

m/ L
MgL
m
Since it might be difficult to measure L precisely, we eliminate L 
so v 
P16.31
T g
2
Mg T g
Tg M
.

m 2
2 m
The total time is the sum of the two times.
In each wire
t
L

L
v
T
Let A represent the cross-sectional area of one wire. The
mass of one wire can be written both as m  V   AL and also as m  L .
Then we have    A 
  d 

 4T 
Thus, t  L 
2
12
 d2
4
12
For copper,
    8 920 1.00  103 2 

t   20.0 
4 150





12
For steel,
    7 860  1.00  103 2 

t   30.0 
4 150





(a)
 0.192 s
0.137  0.192  0.329 s
The total time is
P16.48
 0.137 s
0.175 m   0.350 m  sin  99.6 rad s t 
 sin  99.6 rad s t   0.5
The smallest two angles for which the sine function is 0.5 are 30° and 150°, i.e.,
0.523 6 rad and 2.618 rad.
 99.6 rad s t1  0.523 6 rad , thus t1  5.26 ms
 99.6 rad s t 2  2.618 rad , thus
t 2  26.3 ms
t  t 2  t1  26.3 ms  5.26 ms  21.0 ms
(b)
Distance traveled by the
 99.6 rad s 
 
3
wave    t  
  21.0  10 s  1.68 m .
1.25
rad
m
 k

