 ```P16.2
(a) See ANS FIG P16.2(a); (b) See ANS FIG P16.2(b); (c) The graph (b)
has the same amplitude and wavelength as graph (a). It differs just by being
shifted toward larger x by 2.40 m; (d) The wave has traveled d = vt = 2.40
m to the right.
P16.4
(a) longitudinal P wave; (b) 666 s
P16.6
(a) See ANS FIG P16.6(a); (b) See ANS FIG P16.6(b); (c) See ANS FIG
P16.6(c); (d) See ANS FIG P16.6(d); (e) See ANS FIG P16.6(e)
P16.8
0.800 m/s
P16.10
2.40 m/s
P16.12
 6.67 cm
P16.14
(a) See ANS FIG P16.14; (b) 0.125 s; (c) This agrees with the period found
in the example in the text.
P16.16
(a) 0.100 sin (1.002–20.0t); (b) 3.18 Hz
P16.18
(a) 0.0215 m; (b) 1.95 rad; (c) 5.41 m/s; (d) y(x, t) = (0.0215)sin
(8.38x+80.0+1.95)
P16.20
(a) See ANS FIG P16.20(a); (b) 18.0 rad/m; (c) 0.0833 s; (d) 75.4 rad/s; (e)
4.20 m/s; (f)y 0.200 m sin 18.0x/m 75.4t/s ; (g) y(x, t)=0.200 sin
(18.0x+75.4t0.151), where x and y are in meters and t is in seconds.
P16.22
520 m/s
P16.24
(a) units are newtons; (b) The first T is period of time; the second is force
of tension.
P16.26
(a) y=(2.00104) sin (16.0x 3140t)where y and x are in meters and t is in
seconds; (b) 158 N
P16.28
The calculated gravitational acceleration of the Moon is almost twice that
of the accepted value.
P16.30
(a) v  (30.4) m where v is in meters per second and m is in kilograms;
(b) m = 3.89 kg
P16.32
1.07 kW
P16.34
55.1 Hz
P16.36
(a) As for a string wave, the rate of energy transfer is proportional to the
square of the amplitude to the speed. The rate of energy transfer stays
constant because each wavefront carries constant energy, and the frequency
stays constant. As the speed drops, the amplitude must increase; (b) The
amplitude increases by 5.00 times
P16.38
P16.40
P16.42
See P16.38 for the full explanation.
2P0
(a) A = 40.0; (b) A = 7.00, B = 0, and C = 3.00; (c) In order for two vectors
to be equal, they must have the same magnitude and the same direction in
three-directional space. All of their components must be equal, so all
coefficients of the unit vectors must be equal; (d) A = 0, B = 7.00, C = 3.00,
D = 4.00, E = 2.00; (e) Identify corresponding parts. In order for two
functions to be identically equal, corresponding parts must be identical.
The argument of the sine function must have no units or be equal to units
P16.44
1
(a) See P16.44(a) for full explanation; (b) f ( x  vt )  ( x  vt )2 and
2
1
g ( x  vt )  ( x  vt )2 ;
2
1
1
(c) f ( x  vt )  sin( x  vt ) and g ( x  vt )  sin( x  vt )
2
2
P16.46
1 min
P16.48
(a) 375 m/s2; (b) 0.0450 N; (c) 46.9 N. The maximum transverse force is
very small compared to the tension, more than a thousand times smaller.
2Mg
; (c)
k
2 Mg 
2Mg 
 L0 

k 
k 
P16.50
(a) 2 Mg; (b) L0 
P16.52
(a) The energy a wave crest carries is constant in the absence of absorption.
Then the rate at which energy passes a stationary point, which is the power
of the wave, is constant; (b) The power is proportional to the square of the
amplitude and to the wave speed. The speed decreases as the wave moves
into shallower water near shore, so the amplitude must increase; (c) 8.31
m; (d) As the water depth goes to zero, our model would predict zero speed
and infinite amplitude. In fact, the amplitude must be finite as the wave
comes ashore. As the speed decreases, the wavelength also decreases.
When it becomes comparable to the water depth, or smaller, our formula
gd for wave speed no longer applies.
P16.54
2
(a) P  0.0500 v y ,max ; where P is in watts and vy,max is in meters per second;
(b) The power is proportional to the square of the maximum particle speed;
4
2
(c) E   7.5 10  vy ,max where E is in joules and vy,max is in meters per
second; (d)
1 2
2
mvy ,max ; (e) E  0.300vy ,max where E is in joules and vy,max is
2
in meters per second
P16.56
8.43 &times; 10−3 s
P16.58
2
P16.60
3.86 &times; 10−4
P16.62
This is greater than the average ocean depth, about 4280 m.
P16.64
(a) t  2
P16.66
(a)  ( x) 
L
g
L
g


M  m  M ; (b) 2
L
; (c)
g
mL
Mg
2L
(  L  0 ) x
 L3/ 2  03/ 2 

 0 ; (b) t 
2 T   L  0 
L
```