15
MECHANICAL WAVES
15.1.
IDENTIFY: v  f  . T  1/ f is the time for one complete vibration.
SET UP: The frequency of the note one octave higher is 1568 Hz.
v 344 m/s
1
 0.439 m . T   1.28 ms .
EXECUTE: (a)   
f
784 Hz
f
v 344 m/s

 0.219 m .
f 1568 Hz
EVALUATE: When f is doubled,  is halved.
IDENTIFY: The distance between adjacent dots is  . v  f  . The long-wavelength sound has the lowest
frequency, 20.0 Hz, and the short-wavelength sound has the highest frequency, 20.0 kHz.
SET UP: For sound in air, v  344 m/s .
v 344 m/s
 17.2 m .
EXECUTE: (a) Red dots:   
f 20.0 Hz
(b)  
15.2.
344 m/s
 0.0172 m  1.72 cm .
20.0 103 Hz
(b) In each case the separation easily can be measured with a meterstick.
v 1480 m/s
 74.0 m .
(c) Red dots:   
f
20.0 Hz
Blue dots:  
1480 m/s
 0.0740 m  7.40 cm . In each case the separation easily can be measured with a
20.0 103 Hz
meterstick, although for the red dots a long tape measure would be more convenient.
EVALUATE: Larger wavelengths correspond to smaller frequencies. When the wave speed increases, for a given
frequency, the wavelength increases.
IDENTIFY: v  f    / T .
SET UP: 1.0 h  3600 s . The crest to crest distance is  .
800  103 m
800 km
 220 m/s . v 
EXECUTE: v 
 800 km/h .
3600 s
1.0 h
EVALUATE: Since the wave speed is very high, the wave strikes with very little warning.
IDENTIFY: f  v
SET UP: 1.0 mm  0.0010 m
v 1500 m s
EXECUTE: f  
 1.5 106 Hz
 0.0010 m
EVALUATE: The frequency is much higher than the upper range of human hearing.
IDENTIFY: v  f  . T  1/ f .
Blue dots:  
15.3.
15.4.
15.5.
SET UP:
1 nm  109 m
EXECUTE: (a)   400 nm : f 
c


3.00  108 m/s
 7.50  1014 Hz . T  1/ f  1.33  1015 s .
400  109 m
3.00 10 m/s
 4.29  1014 Hz . T  2.33 1015 s . The frequencies of visible light lie between
700 109 m
4.29 1014 Hz and 7.50 1014 Hz . The periods lie between 1.33 1015 s and 2.33 1015 s .
(b) T is very short and cannot be measured with a stopwatch.
EVALUATE: Longer wavelength corresponds to smaller frequency and larger period.
  700 nm : f 
8
15-1
15-2
15.6.
15.7.
Chapter 15
IDENTIFY: Compare y( x, t ) given in the problem to the general form of Eq.(15.4). f  1/ T and v  f 
SET UP: The comparison gives A  6.50 mm ,   28.0 cm and T  0.0360 s .
EXECUTE: (a) 6.50 mm
(b) 28.0 cm
1
 27.8 Hz
(c) f 
0.0360 s
(d) v  (0.280 m)(27.8 Hz)  7.78 m s
(e) Since there is a minus sign in front of the t / T term, the wave is traveling in the  x-direction .
EVALUATE: The speed of propagation does not depend on the amplitude of the wave.
IDENTIFY: Use Eq.(15.1) to calculate v. T  1/ f and k is defined by Eq.(15.5). The general form of the wave
function is given by Eq.(15.8), which is the equation for the transverse displacement.
SET UP: v  8.00 m/s, A  0.0700 m,   0.320 m
EXECUTE: (a) v  f  so f  v /   (8.00 m/s)/(0.320 m)  25.0 Hz
T  1/ f  1/ 25.0 Hz  0.0400 s
k  2 /   2 rad/0.320 m  19.6 rad/m
(b) For a wave traveling in the  x-direction,
y( x, t )  Acos2 ( x /   t / T ) (Eq.(15.8).)
At x  0, y(0, t )  Acos2 (t / T ), so y  A at t  0. This equation describes the wave specified in the problem.
Substitute in numerical values:
y( x, t )  (0.0700 m)cos(2 ( x / 0.320 m  t / 0.0400 s)).
Or, y ( x, t )  (0.0700 m)cos((19.6 m 1 ) x  (157 rad/s)t ).
(c) From part (b), y  (0.0700 m)cos(2 ( x / 0.320 m  t / 0.0400 s)).
Plug in x  0.360 m and t  0.150 s:
y  (0.0700 m)cos(2 (0.360 m/0.320 m  0.150 s/0.0400 s))
y  (0.0700 m)cos[2 (4.875 rad)]  0.0495 m  4.95 cm
(d) In part (c) t  0.150 s.
y  A means cos(2 ( x /   t / T ))  1
cos  1 for   0, 2 , 4 ,  n(2 ) or n  0, 1, 2,
So y  A when 2 ( x /   t / T )  n(2 ) or x /   x / T  n
15.8.
t  T (n  x / )  (0.0400 s)(n  0.360 m/0.320 m)  (0.0400 s)(n 1.125)
For n  4, t  0.1150 s (before the instant in part (c))
For n  5, t  0.1550 s (the first occurrence of y  A after the instant in part (c)) Thus the elapsed time is
0.1550 s  0.1500 s  0.0050 s.
EVALUATE: Part (d) says y  A at 0.115 s and next at 0.155 s; the difference between these two times is 0.040 s,
which is the period. At t  0.150 s the particle at x  0.360 m is at y  4.95 cm and traveling upward. It takes
T / 4  0.0100 s for it to travel from y  0 to y  A, so our answer of 0.0050 s is reasonable.
IDENTIFY: The general form of the wave function for a wave traveling in the  x-direction is given by Eq.(15.8).
The time for one complete cycle to pass a point is the period T and the number that pass per second is the
frequency f. The speed of a crest is the wave speed v and the maximum speed of a particle in the medium
is vmax   A .
SET UP: Comparison to Eq.(15.8) gives A  3.75 cm , k  0.450 rad/cm and   5.40 rad/s .
2 rad
2 rad
EXECUTE: (a) T 

 1.16 s . In one cycle a wave crest travels a distance

5.40 rad/s
2 rad
2 rad


 0.140 m .
k
0.450 rad/cm
(b) k  0.450 rad/cm . f  1/ T  0.862 Hz  0.862 waves/second .
(c) v  f   (0.862 Hz)(0.140 m)  0.121 m/s . vmax   A  (5.40 rad/s)(3.75 cm)  0.202 m/s .
EVALUATE: The transverse velocity of the particles in the medium (water) is not the same as the velocity of the
wave.
Mechanical Waves
15.9.
15.10.
15-3
IDENTIFY: Evaluate the partial derivatives and see if Eq.(15.12) is satisfied.



SET UP:
cos(kx  t )  k sin(kx  t ) . cos(kx  t )   sin(kx  t ) .
sin(kx  t )  k cos(kx  t ) .
x
t
x

sin(kx  t )   sin(kx  t ) .
t
2 y
2 y
2
EXECUTE: (a)
.


Ak
cos(
kx


t
)
  A 2 cos(kx  t ) . Eq.(15.12) is satisfied, if v   / k .
x 2
t 2
2 y
2 y
2
(b)
.


Ak
sin(
kx


t
)
  A 2 sin(kx  t ) . Eq.(15.12) is satisfied, if v   / k .
x 2
t 2
2 y
2 y
y
y
(c)
 k 2 A cos(kx) .
 kAsin(kx) .
  Asin(t ) . 2   2 A cos(t ) . Eq.(15.12) is not satisfied.
2
x
t
x
t
2
 y
y
(d) vy 
  A cos(kx  t ) . a y  2   A 2 sin(kx  t )
t
t
EVALUATE: The functions cos(kx  t ) and sin(kx  t ) differ only in phase.
IDENTIFY:
v y and a y are given by Eqs.(15.9) and (15.10).
SET UP: The sign of v y determines the direction of motion of a particle on the string. If v y  0 and a y  0 the
speed of the particle is increasing. If v y  0 , the particle is speeding up if v y and a y have the same sign and
slowing down if they have opposite signs.
EXECUTE: (a) The graphs are given in Figure 15.10.
(b) (i) v y  ωA sin(0)  0 and the particle is instantaneously at rest. ay  ω2 A cos(0)  ω2 A and the particle is
speeding up.
(ii) vy  ωAsin(π 4)  ωA
2 , and the particle is moving up. a y  ω2 A cos(π 4)   ω2 A
2 , and the particle is
slowing down ( v y and a y have opposite sign).
(iii) v y  ωA sin(π 2)  ωA and the particle is moving up. ay  ω2 Acos(π 2)  0 and the particle is
instantaneously not accelerating.
(iv) vy  ωAsin(3π 4)  ωA
2 , and the particle is moving up. ay  ω2 A cos(3π 4)  ω2 A
2 , and the particle
is speeding up.
(v) v y  ωA sin(π )  0 and the particle is instantaneously at rest. ay  ω2 A cos(π )  ω2 A and the particle is
speeding up.
(vi) vy  ωAsin(5π 4)  ωA
2 and the particle is moving down. a y  ω2 A cos(5π 4)  ω2 A
2 and the
particle is slowing down ( v y and a y have opposite sign).
(vii) v y  ωA sin(3π 2)  ωA and the particle is moving down. ay  ω2 Acos(3π 2)  0 and the particle is
instantaneously not accelerating.
(viii) vy  ωAsin(7π 4)   ωA
2 , and the particle is moving down. ay  ω2 A cos(7π 4)  ω2 A
2 and the
particle is speeding up ( v y and a y have the same sign).
EVALUATE: At t  0 the wave is represented by Figure 15.10a in the textbook: point (i) in the problem
corresponds to the origin, and points (ii)-(viii) correspond to the points in the figure labeled 1-7. Our results agree
with what is shown in the figure.
Figure 15.10
15-4
Chapter 15
15.11.
IDENTIFY and SET UP: Read A and T from the graph. Apply Eq.(15.4) to determine  and then use Eq.(15.1) to
calculate v.
EXECUTE: (a) The maximum y is 4 mm (read from graph).
(b) For either x the time for one full cycle is 0.040 s; this is the period.
(c) Since y  0 for x  0 and t  0 and since the wave is traveling in the  x-direction then
y( x, t )  Asin[2 (t / T  x / )]. (The phase is different from the wave described by Eq.(15.4); for that wave y  A
for x  0, t  0.) From the graph, if the wave is traveling in the  x-direction and if x  0 and x  0.090 m are
within one wavelength the peak at t  0.01 s for x  0 moves so that it occurs at t  0.035 s (read from graph so
is approximate) for x  0.090 m. The peak for x  0 is the first peak past t  0 so corresponds to the first
maximum in sin[2 (t / T  x / )] and hence occurs at 2 (t / T  x / )   / 2. If this same peak moves to
t1  0.035 s at x1  0.090 m, then
2 (t1 / T  x1 /  )   / 2
Solve for  : t1 / T  x1 /   1/ 4
x1 /   t1 / T  1/ 4  0.035 s / 0.040 s  0.25  0.625
  x1 / 0.625  0.090 m/ 0.625  0.14 m.
Then v  f    / T  0.14 m/ 0.040 s  3.5 m/s.
(d) If the wave is traveling in the  x-direction, then y( x, t )  Asin(2 (t / T  x / )) and the peak at t  0.050 s
for x  0 corresponds to the peak at t1  0.035 s for x1  0.090 m. This peak at x  0 is the second peak past the
origin so corresponds to 2 (t / T  x /  )  5 / 2. If this same peak moves to t1  0.035 s for x1  0.090 m, then
2 (t1 / T  x1 /  )  5 / 2.
t1 / T  x1 /   5/ 4
x1 /   5/ 4  t1 / T  5/ 4  0.035 s / 0.040 s  0.375
  x1 / 0.375  0.090 m/ 0.375  0.24 m.
Then v  f    / T  0.24 m/ 0.040 s  6.0 m/s.
15.12.
EVALUATE: No. Wouldn’t know which point in the wave at x  0 moved to which point at x  0.090 m.
y
IDENTIFY: v y 
. v  f    /T .
t

 2

 2 v   2

SET UP:
A cos  ( x  vt )    A
 sin  ( x  vt ) 
t
 

    


x
t
2


2




EXECUTE: (a) A cos2      A cos  x  t    A cos  x  vt  where   f  v has been used.
T
  T 

 T 
y 2 v
2

Asin  x  vt .
t


(c) The speed is the greatest when the sine is 1, and that speed is 2 vA  . This will be equal to v if A   2 ,
(b) vy 
15.13.
less than v if A   2 and greater than v if A   2 .
EVALUATE: The propagation speed applies to all points on the string. The transverse speed of a particle of the
string depends on both x and t.
IDENTIFY: Follow the procedure specified in the problem.
SET UP: For  and x in cm, v in cm/s and t in s, the argument of the cosine is in radians.
EXECUTE: (a) t  0 :
x (cm)
0.00
1.50
3.00
4.50
6.00
7.50
9.00
10.50
12.00
0.300
0.212
y (cm) 0.300
0.212
0
0.212
0
0.212
0.300
The graph is shown in Figure 15.13a.
(b) (i) t  0.400 s:
x (cm) 0.00
1.50
3.00
4.50
6.00
7.50
9.00
10.50
12.00
y (cm) 0.221 0.0131
0.203
0.300
0.221
0.0131 0.203 0.300 0.221
The graph is shown in Figure 15.13b.
(ii) t  0.800 s:
x (cm)
0.00
1.50
3.00
4.50
6.00
7.50
9.00
10.50
12.00
0.193
0.300 0.230
0.0262
y (cm) 0.0262
0.193
0.300
0.230 0.0262
The graph is shown in Figure 15.13c.
(iii) The graphs show that the wave is traveling in  x-direction.
Mechanical Waves
15-5
EVALUATE: We know that Eq.(15.3) is for a wave traveling in the  x-direction, and y( x, t ) is derived from this.
This is consistent with the direction of propagation we deduced from our graph.
Figure 15.13
15.14.
IDENTIFY: The frequency and wavelength determine the wave speed and the wave speed depends on the
tension.
F
SET UP: v 
.   m /L . v  f  .

0.120 kg
([40.0 Hz][0.750 m])2  43.2 N
2.50 m
EVALUATE: If the frequency is held fixed, increasing the tension will increase the wavelength.
IDENTIFY and SET UP: Use Eq.(15.13) to calculate the wave speed. Then use Eq.(15.1) to calculate the
wavelength.
EXECUTE: (a) The tension F in the rope is the weight of the hanging mass:
F  mg  (1.50 kg)(9.80 m/s 2 )  14.7 N
EXECUTE:
15.15.
F  v 2   ( f  ) 2 
v  F/  14.7 N/(0.0550 kg/m)  16.3 m/s
(b) v  f  so   v / f  (16.3 m/s)/120 Hz  0.136 m.
(c) EVALUATE:
v  F/ , where F  mg. Doubling m increases v by a factor of
and v increases by a factor of
15.16.
2, so  increases by a factor of
2.   v/f . f remains 120 Hz
2.
IDENTIFY: For transverse waves on a string, v  F /  . The general form of the equation for waves traveling in
the  x-direction is y( x, t )  Acos(kx  t ) . For waves traveling in the  x-direction it is y( x, t )  Acos(kx  t ) .
v /k .
15-6
Chapter 15
SET UP: Comparison to the general equation gives A  8.50 mm , k  172 rad/m and   2730 rad/s . The string
has mass 0.128 kg and   m / L  0.0850 kg/m .

2730 rad/s
d 1.50 m

 15.9 m/s . t  
 0.0943 s .
k 172 rad/m
v 15.9 m/s
(b) W  F   v 2  (0.0850 kg/m)(15.9 m/s) 2  21.5 N .
EXECUTE: (a) v 
2 rad
2 rad

 0.0365 m . The number of wavelengths along the length of the string is
k
172 rad/m
1.50 m
 41.1 .
0.0365 m
(d) For a wave traveling in the opposite direction, y( x, t )  (8.50 mm)cos([172 rad/m]x  [2730 rad/s]t)
EVALUATE: We have assumed that the tension in the string is constant and equal to W. In reality the tension will
vary along the length of the string because of the weight of the string and the wave speed will therefore vary along
the string. The tension at the lower end of the string will be W  21.5 N and at the upper end it is
W  1.25 N  22.8 N , an increase of 6%.
(c)  
15.17.
IDENTIFY: For transverse waves on a string, v  F /  . v  f  .
SET UP: The wire has   m / L  (0.0165 kg)/(0.750 m)  0.0220 kg/m .
EXECUTE: (a) v  f   (875 Hz)(3.33  102 m)  29.1 m/s . The tension is
F   v 2  (0.0220 kg/m)(29.1 m/s) 2  18.6 N .
(b) v  29.1 m/s
EVALUATE: If  is kept fixed, the wave speed and the frequency increase when the tension is increased.
15.18.
IDENTIFY: Apply
SET UP:
F
y
 0 to determine the tension at different points of the rope. v  F /  .
From Example 15.3, msamples  20.0 kg , mrope  2.00 kg and   0.0250 kg/m
EXECUTE: (a) The tension at the bottom of the rope is due to the weight of the load, and the speed is the same
88.5m s as found in Example 15.3.
(b) The tension at the middle of the rope is (21.0 kg )(9.80m s 2 )  205.8 N and the wave speed is 90.7 m s.
15.19.
(c) The tension at the top of the rope is (22.0 kg)(9.80 m/s 2 )  215.6 m/s and the speed is 92.9 m/s. (See
Challenge Problem (15.82) for the effects of varying tension on the time it takes to send signals.)
EVALUATE: The tension increases toward the top of the rope, so the wave speed increases from the bottom of the
rope to the top of the rope.
IDENTIFY: v  F /  . v  f  . The general form for y( x, t ) is given in Eq.(15.4), where T  1/ f . Eq.(15.10)
says that the maximum transverse acceleration is amax   2 A  (2 f ) 2 A .
SET UP:   0.0500 kg/m
EXECUTE: (a) v  F   (5.00 N) (0.0500 kg m)  10.0 m s
(b)   v f  (10.0 m s) (40.0 Hz)  0.250 m
(c) y( x, t )  A cos(kx  t ) . k  2   8.00 rad m;   2 f  80.0 rad s .
y( x, t )  (3.00 cm)cos[ (8.00 rad m) x  (80.0 rad s)t ]
(d) vy   A sin(kx  t ) and ay   A 2cos(kx  t ) . ay , max  A 2  A(2 f )2  1890 m s2 .
(e) a y ,max is much larger than g, so it is a reasonable approximation to ignore gravity.
15.20.
EVALUATE: y( x, t ) in part (c) gives y(0,0)  A , which does correspond to the oscillator having maximum
upward displacement at t  0 .
IDENTIFY: Apply Eq.(15.25).
SET UP:   2 f .   m / L .
1  3.00  103 kg 
1
2
3
2
 F  2 A2 . Pav 

 (25.0 N)(2 (120.0 Hz)) (1.6 10 m)  0.223 W
2 
0.80 m
2

or 0.22 W to two figures.
(b) Pav is proportional to A2 , so halving the amplitude quarters the average power, to 0.056 W.
EVALUATE: The average power is also proportional to the square of the frequency.
EXECUTE: (a) Pav 
Mechanical Waves
15.21.
IDENTIFY:
SET UP:
For a point source, I 
15-7
P
I
r2
and 1  22 .
2
I 2 r1
4 r
1  W  106 W
EXECUTE: (a) r2  r1
I1
10.0 W/m2
 (30.0 m)
 95 km
I2
1106 W/m2
2
(b)
15.22.
r 
I 2 r32
 2 , with I 2  1.0  W/m 2 and r3  2r2 . I 3  I 2  2   I 2 / 4  0.25  W/m2 .
I 3 r2
 r3 
(c) P  I (4 r 2 )  (10.0 W/m 2 )(4 )(30.0 m) 2  1.1  105 W
EVALUATE: These are approximate calculations, that assume the sound is emitted uniformly in all directions and
that ignore the effects of reflection, for example reflections from the ground.
IDENTIFY: Apply Eq.(15.26).
SET UP: I1  0.11 W/m 2 . r1  7.5 m . Set I 2  1.0 W/m 2 and solve for r2 .
EXECUTE:
15.23.
r2  r1
I1
0.11 W m2
 (7.5 m)
 2.5 m, so it is possible to move
I2
1.0 W m2
r1  r2  7.5 m  2.5 m  5.0 m closer to the source.
EVALUATE: I increases as the distance r of the observer from the source decreases.
IDENTIFY and SET UP: Apply Eq.(15.26) to relate I and r.
Power is related to intensity at a distance r by P  I (4 r 2 ). Energy is power times time.
EXECUTE:
(a) I1r12  I 2 r22
I 2  I1 (r1 / r2 ) 2  (0.026 W/m 2 )(4.3 m/3.1 m) 2  0.050 W/m 2
(b) P  4 r 2 I  4 (4.3 m) 2 (0.026 W/m 2 )  6.04 W
Energy  Pt  (6.04 W)(3600 s)  2.2  104 J
15.24.
EVALUATE: We could have used r  3.1 m and I  0.050 W/m2 in P  4 r 2 I and would have obtained the
same P. Intensity becomes less as r increases because the radiated power spreads over a sphere of larger area.
IDENTIFY: The tension and mass per unit length of the rope determine the wave speed. Compare y( x, t ) given in
the problem to the general form given in Eq.(15.8). v   / k . The average power is given by Eq. (15.25).
SET UP: Comparison with Eq.(15.8) gives A  2.33 mm , k  6.98 rad/m and   742 rad/s .
EXECUTE: (a) A  2.30 mm
 742 rad s
(b) f 

 118 Hz .
2
2
2
(c)   2 
 0.90 m
k
6.98 rad m
742 rad s
(d) v   
 106 m s
k 6.98 rad m
(e) The wave is traveling in the  x direction because the phase of y( x, t ) has the form kx  ωt.
(f ) The linear mass density is μ  (3.38  103 kg) (1.35 m)  2.504 10 3 kg/m , so the tension is
F   v 2  (2.504  103 kg m)(106.3 m s) 2  28.3 N .
(g) Pav  12  F  2 A2  12 (2.50 103 kg m)(28.3 N)(742 rad s) 2 (2.30 103 m) 2  0.39 W
EVALUATE: In part (d) we could also calculate the wave speed as v  f  and we would obtain the same result.
15.25.
IDENTIFY: P  4 r 2 I
SET UP: From Example 15.5, I  0.250 W/m2 at r  15.0 m
EXECUTE: P  4 r 2 I  4 (15.0 m) 2 (0.250 W/m 2 )  707 W
EVALUATE: I  0.010 W/m2 at 75.0 m and 4 (75.0 m) 2 (0.010 W/m 2 )  707 W . P is the average power of the
sinusoidal waves emitted by the source.
15-8
Chapter 15
15.26.
IDENTIFY: The distance the wave shape travels in time t is vt. The wave pulse reflects at the end of the string, at
point O.
SET UP: The reflected pulse is inverted when O is a fixed end and is not inverted when O is a free end.
EXECUTE: (a) The wave form for the given times, respectively, is shown in Figure 15.26a.
(b) The wave form for the given times, respectively, is shown in Figure 15.26b.
EVALUATE: For the fixed end the result of the reflection is an inverted pulse traveling to the left and for the free
end the result is an upright pulse traveling to the left.
Figure 15.26
15.27.
IDENTIFY: The distance the wave shape travels in time t is vt. The wave pulse reflects at the end of the string, at
point O.
SET UP: The reflected pulse is inverted when O is a fixed end and is not inverted when O is a free end.
EXECUTE: (a) The wave form for the given times, respectively, is shown in Figure 15.27a.
(b) The wave form for the given times, respectively, is shown in Figure 15.27b.
EVALUATE: For the fixed end the result of the reflection is an inverted pulse traveling to the left and for the free
end the result is an upright pulse traveling to the left.
Figure 15.27
15.28.
IDENTIFY: Apply the principle of superposition.
SET UP: The net displacement is the algebraic sum of the displacements due to each pulse.
EXECUTE: The shape of the string at each specified time is shown in Figure 15.28.
EVALUATE: The pulses interfere when they overlap but resume their original shape after they have completely
passed through each other.
Figure 15.28
15.29.
IDENTIFY: Apply the principle of superposition.
SET UP: The net displacement is the algebraic sum of the displacements due to each pulse.
EXECUTE: The shape of the string at each specified time is shown in Figure 15.29.
Mechanical Waves
15-9
EVALUATE: The pulses interfere when they overlap but resume their original shape after they have completely
passed through each other.
Figure 15.29
15-10
Chapter 15
15.30.
IDENTIFY: Apply the principle of superposition.
SET UP: The net displacement is the algebraic sum of the displacements due to each pulse.
EXECUTE: The shape of the string at each specified time is shown in Figure 15.30.
EVALUATE: The pulses interfere when they overlap but resume their original shape after they have completely
passed through each other.
Figure 15.30
15.31.
IDENTIFY: Apply the principle of superposition.
SET UP: The net displacement is the algebraic sum of the displacements due to each pulse.
EXECUTE: The shape of the string at each specified time is shown in Figure 15.31.
EVALUATE: The pulses interfere when they overlap but resume their original shape after they have completely
passed through each other.
Figure 15.31
Mechanical Waves
15.32.
15-11
IDENTIFY: ynet  y1  y2 . The string never moves at values of x for which sin kx  0 .
SET UP: sin( A  B)  sin Acos B  cos Asin B
EXECUTE: (a) ynet  Asin(kx  t )  Asin(kx  t ) .
ynet  A[sin(kx)cos(t )  cos(kx)sin(t )  sin(kx)cos(t )  cos(kx)sin(t )]  2 Asin(kx)cos(t )
n
n
n
.


k 2 /  2
EVALUATE: Using y  Asin(kx  t ) instead of y  Acos(kx  t ) corresponds to a particular choice of phase
and corresponds to y  0 at x  0 , for all t.
IDENTIFY and SET UP: Nodes occur where sin kx  0 and antinodes are where sin kx  1.
EXECUTE: Eq.(15.28): y  ( ASW sin kx)sin t
(a) At a node y  0 for all t. This requires that sin kx  0 and this occurs for kx  n , n  0, 1, 2,
(b) sin kx  0 for kx  n , n  0, 1, 2,
15.33.
. x
n
 (1.33 m)n, n  0, 1, 2,
0.750 rad/m
(b) At an antinode sin kx  1 so y will have maximum amplitude. This occurs when kx   n  12  ,
x  n / k 
n  0, 1, 2,

 (1.33 m)  n  12  , n  0, 1, 2,
0.750 rad/m
EVALUATE:   2 / k  2.66 m. Adjacent nodes are separated by  / 2, adjacent antinodes are separated by
 / 2, and the node to antinode distance is  / 4.
IDENTIFY: Apply Eqs.(15.28) and (15.1). At an antinode, y(t )  ASW sin t . k and  for the standing wave have
x   n  12  / k   n  12 
15.34.
the same values as for the two traveling waves.
SET UP: ASW  0.850 cm . The antinode to antinode distance is  / 2 , so   30.0 cm . v y  y / t .
EXECUTE: (a) The node to node distance is  / 2  15.0 cm .
(b)  is the same as for the standing wave, so   30.0 cm . A  12 ASW  0.425 cm .
v f 


0.300 m
 13.3 m/s .
0.0750 s
T
y
(c) vy 
 ASW sin kx cos t . At an antinode sin kx  1 , so v y  ASW cos t . vmax  ASW .
t
2 rad 2 rad


 83.8 rad/s . vmax  (0.850  102 m)(83.8 rad/s)  0.0712 m/s . vmin  0 .
T
0.0750 s
(d) The distance from a node to an adjacent antinode is  / 4  7.50 cm .
EVALUATE: The maximum transverse speed for a point at an antinode of the standing wave is twice the
maximum transverse speed for each traveling wave, since ASW  2 A .
15.35.
IDENTIFY: Evaluate  2 y / x 2 and  2 y / t 2 and see if Eq.(15.12) is satisfied for v   / k .




sin kx  k cos kx .
cos kx  k sin kx . sin t   cost . cost   sin t
t
x
x
t
2
2 y

y
 k 2  Asw sin t sin kx,
  2  Asw sin t sin kx, so for y( x, t ) to be a solution of
EXECUTE: (a)
x 2
t 2
 2

Eq.(15.12), k 2  2 , and v  .
v
k
(b) A standing wave is built up by the superposition of traveling waves, to which the relationship v   k applies.
SET UP:
15.36.
EVALUATE: y( x, t )  ( ASW sin kx)sin t is a solution of the wave equation because it is a sum of solutions to the
wave equation.
IDENTIFY and SET UP: cos(kx  t )  cos kx cost sin kx sin t
EXECUTE:
y1  y2  A [ cos(kx  ωt )  cos(kx  ωt )] .
y1  y2  A[ cos kx cos t  sin kx sin t  cos kx cost  sin kx sin t ]  2 A sin kx sin t .
EVALUATE: The derivation shows that the standing wave of Eq.(15.28) results from the combination of two
waves with the same A, k, and  that are traveling in opposite directions.
15-12
Chapter 15
15.37.
IDENTIFY: Evaluate  2 y / x 2 and  2 y / t 2 and show that Eq.(15.12) is satisfied.

y y

y y
( y1  y2 )  1  2 and ( y1  y2 )  1  2
x
x x
t
t
t
 2 y  2 y1  2 y2
 2 y  2 y1  2 y2
EXECUTE:

 2 and 2  2  2 . The functions y1 and y2 are given as being solutions to
x 2 x 2
x
t
t
t
the wave equation, so
 2 y  2 y1  2 y2  1   2 y1  1   2 y2  1    2 y1  2 y2   1   2 y

 2   2  2   2  2   2   2  2    2  2 and so y  y1  y2 is a solution of
x 2 x 2
x
t   v  t
 v  t
 v  t
 v   t
Eq. (15.12).
EVALUATE: The wave equation is a linear equation, as it is linear in the derivatives, and differentiation is a linear
operation.
2L
 v 
IDENTIFY: For a string fixed at both ends, n 
and f n  n   .
n
 2L 
SET UP: For the fundamental, n  1 . For the second overtone, n  3 . For the fourth harmonic, n  4 .
(48.0 m s)
EXECUTE: (a) 1  2L  3.00 m . f1  v 
 16.0 Hz .
2L 2(1.50 m)
SET UP:
15.38.
(b) 3  1 3  1.00 m . f 2  3 f1  48.0 Hz .
15.39.
(c) 4  1 4  0.75 m . f3  4 f1  64.0 Hz .
EVALUATE: As n increases,  decreases and f increases.
IDENTIFY: Use Eq.(15.1) for v and Eq.(15.13) for the tension F. v y  y / t and a y  v y / t.
(a) SET UP: The fundamental standing wave is sketched in Figure 15.39.
f  60.0 Hz
From the sketch,
 / 2  L so
  2L  1.60 m
Figure 15.39
EXECUTE: v  f   (60.0 Hz)(1.60 m)  96.0 m/s
(b) The tension is related to the wave speed by Eq.(15.13):
v  F /  so F  v 2 .
  m / L  0.0400 kg/0.800 m  0.0500 kg/m
F   v 2  (0.0500 kg/m)(96.0 m/s) 2  461 N.
(c)   2 f  377 rad/s and y( x, t )  ASW sin kx sin t
v y   ASW sin kx cos t ; ay   2 ASW sin kx sin t
(v y ) max   ASW  (377 rad/s)(0.300 cm)  1.13 m/s.
15.40.
(ay )max   2 ASW  (377 rad/s)2 (0.300 cm)  426 m/s2 .
EVALUATE: The transverse velocity is different from the wave velocity. The wave velocity and tension are
similar in magnitude to the values in the Examples in the text. Note that the transverse acceleration is quite large.
IDENTIFY: The fundamental frequency depends on the wave speed, and that in turn depends on the tension.
F
v
SET UP: v 
where   m / L . f1 
. The nth harmonic has frequency f n  nf1 .

2L
EXECUTE: (a) v 
F

m/ L
FL
(800 N)(0.400 m)
v
327 m/s

 327 m/s . f1 

 409 Hz .
3
m
3.00  10 kg
2 L 2(0.400 m)
10,000 Hz
 24.4 . The 24th harmonic is the highest that could be heard.
f1
EVALUATE: In part (b) we use the fact that a standing wave on the wire produces a sound wave in air of the same
frequency.
(b) n 
Mechanical Waves
15.41.
15-13
IDENTIFY: Compare y( x, t ) given in the problem to Eq.(15.28). From the frequency and wavelength for the
third harmonic find these values for the eighth harmonic.
(a) SET UP: The third harmonic standing wave pattern is sketched in Figure 15.41.
Figure 15.41
EXECUTE: (b) Eq. (15.28) gives the general equation for a standing wave on a string:
y( x, t )  ( ASW sin kx)sin t
ASW  2 A, so A  ASW / 2  (5.60 cm)/2  2.80 cm
(c) The sketch in part (a) shows that L  3( / 2). k  2 /  ,   2 / k
Comparison of y( x, t ) given in the problem to Eq. (15.28) gives k  0.0340 rad/cm. So,
  2 /(0.0340 rad/cm)  184.8 cm
L  3( / 2)  277 cm
(d)   185 cm, from part (c)
  50.0 rad/s so f   / 2  7.96 Hz
period T  1/ f  0.126 s
v  f   1470 cm/s
(e) v y  dy / dt   ASW sin kx cos t
v y , max   ASW  (50.0 rad/s)(5.60 cm)  280 cm/s
(f ) f3  7.96 Hz  3 f1, so f1  2.65 Hz is the fundamental
15.42.
f8  8 f1  21.2 Hz; 8  2 f8  133 rad/s
  v / f  (1470 cm/s)/(21.2 Hz)  69.3 cm and k  2 /   0.0906 rad/cm
y( x, t )  (5.60 cm)sin([0.0906 rad/cm]x)sin([133 rad/s]t )
EVALUATE: The wavelength and frequency of the standing wave equals the wavelength and frequency of the two
traveling waves that combine to form the standing wave. In the 8th harmonic the frequency and wave number are
larger than in the 3rd harmonic.
IDENTIFY: Compare the y( x, t ) specified in the problem to the general form of Eq.(15.28).
SET UP: The comparison gives ASW  4.44 mm , k  32.5 rad/m and   754 rad/s .
EXECUTE: (a) A  12 ASW  12 (4.44 mm)  2.22 mm .
2
(b)   2 
 0.193 m.
k
32.5 rad m
754 rad s
 120 Hz .
(c) f   
2
2
754 rad s
(d) v   
 23.2 m s.
k 32.5 rad m
(e) If the wave traveling in the  x direction is written as y1 ( x, t )  A cos(kx  t ), then the wave traveling in the
15.43.
 x-direction is y2 ( x, t )   A cos(kx  t ) , where A  2.22 mm from part (a), k  32.5 rad m and   754 rad s .
(f ) The harmonic cannot be determined because the length of the string is not specified.
EVALUATE: The two traveling waves that produce the standing wave are identical except for their direction of
propagation.
(a) IDENTIFY and SET UP: Use the angular frequency and wave number for the traveling waves in Eq.(15.28) for
the standing wave.
EXECUTE: The traveling wave is y( x, t )  (2.30 mm)cos([6.98 rad/m]x)  [742 rad/s]t )
A  2.30 mm so ASW  4.60 mm; k  6.98 rad/m and   742 rad/s
The general equation for a standing wave is y( x, t )  ( ASW sin kx)sin t , so
y( x, t )  (4.60 mm)sin([6.98 rad/m]x)sin([742 rad/s]t )
(b) IDENTIFY and SET UP: Compare the wavelength to the length of the rope in order to identify the harmonic.
EXECUTE: L  1.35 m (from Exercise 15.24)
  2 / k  0.900 m
L  3( / 2), so this is the 3rd harmonic
15-14
Chapter 15
(c) For this 3rd harmonic, f   / 2  118 Hz
f3  3 f1 so f1  (118 Hz)/3  39.3 Hz
EVALUATE: The wavelength and frequency of the standing wave equals the wavelength and frequency of the two
traveling waves that combine to form the standing wave. The nth harmonic has n node-to-node segments and the
node-to-node distance is  / 2, so the relation between L and  for the nth harmonic is L  n(
15.44.
2L
and frequencies f n  nf1 . The
n
standing wave on the string and the sound wave it produces have the same frequency.
SET UP: For the fundamental n  1 and for the second overtone n  3 . The string has
  m / L  (8.75  103 kg) /(0.750 m)  1.17 10 2 kg/m .
EXECUTE: (a)   2L /3  2(0.750 m)/3  0.500 m . The sound wave has frequency
IDENTIFY:
344 m/s
 1.03 104 Hz . For waves on the string,
 3.35 102 m
v  f   (1.03  104 Hz)(0.500 m)  5.15  103 m/s . The tension in the string is
f 
v
v  F /  . v  f  . The standing waves have wavelengths n 

F   v 2  (1.17  102 kg/m)(5.15  103 m/s) 2  3.10 105 N .
15.45.
(b) f1  f3 / 3  (1.03  104 Hz) / 3  3.43 103 Hz .
EVALUATE: The waves on the string have a much longer wavelength than the sound waves in the air because the
speed of the waves on the string is much greater than the speed of sound in air.
IDENTIFY and SET UP: Use the information given about the A 4 note to find the wave speed, that depends on the
linear mass density of the string and the tension. The wave speed isn’t affected by the placement of the fingers on the
bridge. Then find the wavelength for the D 5 note and relate this to the length of the vibrating portion of the string.
EXECUTE: (a) f  440 Hz when a length L  0.600 m vibrates; use this information to calculate the speed v
of waves on the string. For the fundamental  / 2  L so   2L  2(0.600 m)  1.20 m. Then
v  f   (440 Hz)(1.20 m)  528 m/s. Now find the length L  x of the string that makes f  587 Hz.
v 528 m/s

 0.900 m
f 587 Hz
L   /2  0.450 m, so x  0.450 m  45.0 cm.
(b) No retuning means same wave speed as in part (a). Find the length of vibrating string needed to produce
f  392 Hz.

v 528 m/s

 1.35 m
f 392 Hz
L   / 2  0.675 m; string is shorter than this. No, not possible.
EVALUATE: Shortening the length of this vibrating string increases the frequency of the fundamental.
IDENTIFY: y( x, t )  ( ASW sin kx)sin t . v y  y / t . ay  2 y / t 2 .

15.46.
SET UP:
vmax  ( ASW sin kx) . amax  ( ASW sin kx) 2 .
EXECUTE: (a) (i) x   is a node, and there is no motion. (ii) x   is an antinode, and vmax  A(2 f )  2 fA ,
2
4
2 2

1
amax  (2 f )vmax  4 f A . (iii) cos 
and this factor multiplies the results of (ii), so vmax  2 fA ,
4
2
amax  2 2 2 f 2 A .
15.47.
(b) The amplitude is 2 Asin kx, or (i) 0, (ii) 2A, (iii) 2A 2.
(c) The time between the extremes of the motion is the same for any point on the string (although the period of the
zero motion at a node might be considered indeterminate) and is 1/ 2 f .
EVALUATE: Any point in a standing wave moves in SHM. All points move with the same frequency but have
different amplitude.
v
IDENTIFY: For the fundamental, f1 
. v  F /  . A standing wave on a string with frequency f produces a
2L
sound wave that also has frequency f.
SET UP: f1  245 Hz . L  0.635 m .
EXECUTE: (a) v  2 f1L  2(245 Hz)(0.635 m)  311 m s.
Mechanical Waves
15-15
(b) The frequency of the fundamental mode is proportional to the speed and hence to the square root of the tension;
(245 Hz) 1.01  246 Hz .
15.48.
(c) The frequency will be the same, 245 Hz. The wavelength will be air  vair f  (344 m s) (245z)  1.40 m,
which is larger than the wavelength of standing wave on the string by a factor of the ratio of the speeds.
EVALUATE: Increasing the tension increases the wave speed and this in turn increases the frequencies of the
standing waves. The wavelength of each normal mode depends only on the length of the string and doesn't change
when the tension changes.
IDENTIFY: The ends of the stick are free, so they must be displacement antinodes. The first harmonic has one
node, at the center of the stick, and each successive harmonic adds one node.
SET UP: The node to node and antinode to antinode distance is  / 2 .
EXECUTE: The standing wave patterns for the first three harmonics are shown in Figure 15.48.
1
1st harmonic: L  1  1  2L  4.0 m . 2nd harmonic: L  12  2  L  2.0 m .
2
3
2L
rd
3 harmonic: L  3  3 
 1.33 m .
2
3
EVALUATE: The higher the harmonic the shorter the wavelength.
Figure 15.48
15.49.
IDENTIFY and SET UP: Calculate v, , and k from Eqs.(15.1), (15.5), and (15.6). Then apply Eq.(15.7) to obtain
y( x, t ).
A  2.50 103 m,   1.80 m, v  36.0 m/s
EXECUTE: (a) v  f  so f  v /   (36.0 m/s)/1.80 m  20.0 Hz
  2 f  2 (20.0 Hz)  126 rad/s
k  2 /  2 rad/1.80 m  3.49 rad/m
(b) For a wave traveling to the right, y( x, t )  Acos(kx  t ). This equation gives that the x  0 end of the string
has maximum upward displacement at t  0.
Put in the numbers: y ( x, t )  (2.50  103 m)cos((3.49 rad/m)x  (126 rad/s)t ).
(c) The left hand end is located at x  0. Put this value into the equation of part (b):
y (0, t )  (2.50 103 m)cos((126 rad/s)t ).
(d) Put x  1.35 m into the equation of part (b):
y (1.35 m, t )  (2.50  103 m)cos((3.49 rad/m)(1.35 m)  (126 rad/s)t ).
y (1.35 m, t )  (2.50  103 m)cos(4.71 rad)  (126 rad/s)t
4.71 rad  3 / 2 and cos( )  cos( ), so y (1.35 m, t )  (2.50 103 m)cos((126 rad/s)t  3 / 2 rad)
(e) y  Acos(kx  t ) ((part b))
y

 A cos(kx  t )   A sin(kx  t ).
t
t
The maximum v y is A  (2.50  103 m)(126 rad/s)  0.315 m/s.
The transverse velocity is given by vy 
(f ) y ( x, t )  (2.50  103 m)cos((3.49 rad/m)x  (126 rad/s)t )
t  0.0625 s and x  1.35 m gives
y  (2.50  103 m)cos((3.49 rad/m)(1.35 m)  (126 rad/s)(0.0625 s))  2.50  103 m.
v y   A sin(kx  t )  (0.315 m/s)sin((3.49 rad/m)x  (126 rad/s)t )
t  0.0625 s and x  1.35 m gives
v y  (0.315 m/s)sin((3.49 rad/m)(1.35 m)  (126 rad/s)(0.0625 s))  0.0
EVALUATE: The results of part (f ) illustrate that v y  0 when y   A, as we saw from SHM in Chapter 13.
15.50.
IDENTIFY: Compare y( x, t ) given in the problem to the general form given in Eq.(15.8).
SET UP: The comparison gives A  0.750 cm , k  0.400 rad/cm and   250 rad/s .
15-16
Chapter 15
EXECUTE: (a) A  0.750 cm,  
2
 5.00 cm, f  125 Hz, T  1  0.00800 s and
0.400 rad/cm
f
v   f  6.25 m/s.
(b) The sketches of the shape of the rope at each time are given in Figure 15.50.
(c) To stay with a wavefront as t increases, x decreases and so the wave is moving in the  x -direction.
(d) From Eq. (15.13), the tension is F   v 2  (0.50 kg m) (6.25 m s) 2  19.5 N .
(e) Pav  12  F 2 A2  54.2 W.
EVALUATE: The argument of the cosine is (kx  t ) for a wave traveling in the  x-direction , and that is the case
here.
Figure 15.50
15.51.
IDENTIFY: The speed in each segment is v  F μ. The time to travel through a segment is t  L/v.
SET UP: The travel times for each segment are t1  L
EXECUTE: Adding the travel times gives ttotal  L
1
1
F
, t2  L

41

, and t3  L 1 .
F
4F


 2L 1  12 L 1  72 L 1 .
F
F
F
F
(b) No. The speed in a segment depends only on F and  for that segment.
EVALUATE: The wave speed is greater and its travel time smaller when the mass per unit length of the segment
decreases.
Mechanical Waves
15.52.
IDENTIFY: Apply

z
15-17
 0 to find the tension in each wire. Use v  F/ to calculate the wave speed for each
wire and then t  L/v is the time for each pulse to reach the ceiling, where L  1.25 m .
m
2.50 N
 0.204 kg/m . The free-body diagram for the beam is
SET UP: The wires have   
L (9.80 m/s 2 )(1.25 m)
given in Figure 15.52. Take the axis to be at the end of the beam where wire A is attached.
EXECUTE:  z  0 gives TB L  w( L /3) and TB  w /3  583 N . TA  TB  1750 N , so TA  1167 N .
TA
vA 


1167 N
583 N
1.25 m
 75.6 m/s . t A 
 53.5 m/s .
 0.0165 s . vB 
0.204 kg/m
0.204 kg/m
75.6 m/s
1.25 m
 0.0234 s . t  tB  t A  6.9 ms .
53.5 m/s
EVALUATE: The wave pulse travels faster in wire A, since that wire has the greater tension.
tB 
Figure 15.52
15.53.
IDENTIFY and SET UP: The transverse speed of a point of the rope is v y  y/t where y( x, t ) is given by Eq.(15.7).
EXECUTE:
(a) y( x, t )  Acos(kx  t )
v y  dy / dt   A sin(kx  t )
v y , max  A  2 fA
f 
v

and v 
F
 1  FL
, so f   
(m / L)
 M
 2 A  FL
v y , max  

   M
(b) To double v y , max increase F by a factor of 4.
15.54.
EVALUATE: Increasing the tension increases the wave speed v which in turn increases the oscillation frequency.
With the amplitude held fixed, increasing the number of oscillations per second increases the transverse velocity.
IDENTIFY: The maximum vertical acceleration must be at least g .
SET UP:
g 2 
.
4 2 F
EVALUATE: When the amplitude of the motion increases, the maximum acceleration of a point on the rope
increases.
IDENTIFY and SET UP: Use Eq.(15.1) and   2 f to replace v by  in Eq.(15.13). Compare this equation to
EXECUTE:
15.55.
amax   2 A
g   2 Amin and thus Amin  g / 2. Using   2 f  2 v/ and v  F/ , this becomes Amin 
  k  / m from Chapter 13 to deduce k .
EXECUTE: (a)   2 f , f  v/ , and v  F/ These equations combine to give
  2 f  2 (v/ )  (2 /) F/ .
But also   k  / m. Equating these expressions for  gives k   m(2 / ) 2 ( F/ )
But m   x so k   x (2 / ) 2 F
(b) EVALUATE: The “force constant” k  is independent of the amplitude A and mass per unit length  , just as is
the case for a simple harmonic oscillator. The force constant is proportional to the tension in the string F and
inversely proportional to the wavelength  . The tension supplies the restoring force and the 1/  2 factor represents
the dependence of the restoring force on the curvature of the string.
15-18
Chapter 15
15.56.
IDENTIFY: Apply

z
 0 to one post and calculate the tension in the wire. v  F /  for waves on the wire.
v  f  . The standing wave on the wire and the sound it produces have the same frequency. For standing waves on
2L
.
n
SET UP: For the 7th overtone, n  8 . The wire has   m / L  (0.732 kg)/(5.00 m)  0.146 kg/m . The free-body
diagram for one of the posts is given in Figure 15.56. Forces at the pivot aren’t shown. We take the rotation axis to
be at the pivot, so forces at the pivot produce no torque.
w
235 N
L

EXECUTE:  z  0 gives w cos57.0°   T ( L sin57.0°)  0 . T 

 76.3 N . For
2tan57.0° 2tan57.0°
2

the wire, n 
waves on the wire, v 
F


76.3 N
 22.9 m/s . For the 7th overtone standing wave on the wire,
0.146 kg/m
2L 2(5.00 m)
v 22.9 m/s
 18.3 Hz . The sound waves have frequency 18.3 Hz and

 1.25 m . f  
 1.25 m
8
8
344 m/s
wavelength  
 18.8 m
18.3 Hz
EVALUATE: The frequency of the sound wave is at the lower limit of audible frequencies. The wavelength of the
standing wave on the wire is much less than the wavelength of the sound waves, because the speed of the waves on
the wire is much less than the speed of sound in air.

Figure 15.56
15.57.
IDENTIFY: The magnitude of the transverse velocity is related to the slope of the t versus x curve. The transverse
acceleration is related to the curvature of the graph, to the rate at which the slope is changing.
SET UP: If y increases as t increases, v y is positive. a y has the same sign as v y if the transverse speed is
increasing.
EXECUTE: (a) and (b) (1): The curve appears to be horizontal, and v y  0 . As the wave moves, the point will
begin to move downward, and a y  0 . (2): As the wave moves in the  x-direction, the particle will move upward
so v y  0 . The portion of the curve to the left of the point is steeper, so a y  0 . (3) The point is moving down, and
will increase its speed as the wave moves; v y  0 , a y  0 . (4) The curve appears to be horizontal, and v y  0 . As
the wave moves, the point will move away from the x-axis, and a y  0 . (5) The point is moving downward, and
will increase its speed as the wave moves; v y  0, a y < 0 . (6) The particle is moving upward, but the curve that
represents the wave appears to have no curvature, so v y  0 and a y  0 .
(c) The accelerations, which are related to the curvatures, will not change. The transverse velocities will all change
sign.
EVALUATE: At points 1, 3, and 5 the graph has negative curvature and a y  0 . At points 2 and 4 the graph has
positive curvature and a y  0 .
15.58.
IDENTIFY: The time it takes the wave to travel a given distance is determined by the wave speed v. A point on
the string travels a distance 4A in time T.
SET UP: v  f  . T  1/ f .
EXECUTE: (a) The wave travels a horizontal distance d in a time t 
d
d
8.00 m


 0.333 s.
v  f  0.600 m  40.0 Hz 
Mechanical Waves
15-19
(b) A point on the string will travel a vertical distance of 4A each cycle. Although the transverse velocity v y  x, t 
15.59.
is not constant, a distance of h  8.00 m corresponds to a whole number of cycles,
n  h (4 A)  (8.00 m) [4(5.00 103 m)]  400, so the amount of time is t  nT  n f  (400) (40.0 Hz)  10.0 s.
EVALUATE: (c) The time in part (a) is independent of amplitude but the time in part (b) depends on the amplitude
of the wave. For (b), the time is halved if the amplitude is doubled.
IDENTIFY: y 2 ( x, y )  z 2 ( x, y )  A2 . The trajectory is a circle of radius A.
SET UP:
EXECUTE:
v y  y / t , vz  z / t . a y  v y / t , az  vz / t
At t  0, y(0,0)  A, z(0,0)  0. At t   2 , y(0, 2)  0, z(0,  2)   A .
At t    , y(0,  )   A, z(0,  )  0. At t  3 2 , y(0,3 2)  0, z(0,3 2)  A . The trajectory and
these points are sketched in Figure 15.59.
(b) v y  y t   A sin(kx  t ), vz  z t   A cos(kx  t ) .
v = vy ˆj + vz kˆ = A[sin(kx   ) ˆj  cos(kx  t )kˆ ] . v  vy2  vz2  A so the speed is constant.
r  yˆj  zkˆ . r  v  yvy  zvz  A2 sin(kx  t )cos(kx  t )  A2 cos(kx  t ) sin(kx  t )  0 .
r  v  0, so v is tangent to the circular path.
(c) ay  vy t   A 2cos(kx  t ), az vz t   A 2 sin(kx  t )
r  a  yay  zaz   A2 2[cos2 (kx  t )  sin 2 (kx  t )]   A2 2 . r  A , r  a  ra .
r  a  ra cos  so   180 and a is opposite in direction to r ; a is radially inward. For these y( x, t ) and z( x, t ),
y 2  z 2  A2 , so the path is again circular, but the particle rotates in the opposite sense compared to part (a).
EVALUATE: The wave propagates in the  x-direction. The displacement is transverse, so v and a lie in the yzplane.
Figure 15.59
15.60.
IDENTIFY: The wavelengths of the standing waves on the wire are given by n 
2L
. When the ball is changed
n
Fl0
.
AY
SET UP: For the third harmonic, n  3 . For copper, Y  111010 Pa . The wire has cross-sectional area
A   r 2   (0.512  103 m) 2  8.24  107 m 2
the wavelength changes because the length of the wire changes; l 
EXECUTE: (a) 3 
2(1.20 m)
 0.800 m
3
(b) The increase in length when the 100.0 N ball is replaced by the 500.0 N ball is given by l 
(F )l0
, where
AY
F  400.0 N is the increase in the force applied to the end of the wire.
(400.0 N)(1.20 m)
l 
 5.30  103 m . The change in wavelength is   32 l  3.5 mm .
(8.24  107 m 2 )(11  1010 Pa)
EVALUATE: The change in tension changes the wave speed and that in turn changes the frequency of the standing
wave, but the problem asks only about the wavelength.
15-20
Chapter 15
15.61.
IDENTIFY:
Follow the procedure specified in part (b).
u
u
SET UP: If u  x  vt , then
 v and
 1.
t
x
EXECUTE: (a) As time goes on, someone moving with the wave would need to move in such a way that the wave
appears to have the same shape. If this motion can be described by x  vt  b, with b a constant, then
y( x, t )  f (b), and the waveform is the same to such an observer.
(b)
2
2 y d 2 f
2 y
2 d f


v
, so y(x,t )  f ( x  vt ) is a solution to the wave equation with wave speed v .
and
x 2 du 2
t 2
du 2
(c) This is of the form y( x, t )  f (u), with u  x  vt and f (u)  De B
15.62.
2
( x Ct / B )2
. The result of part (b) may be used
to determine the speed v  C B .
EVALUATE: The wave in part (c) moves in the  x-direction. The speed of the wave is independent of the
constant D.
IDENTIFY: The phase angle determines the value of y for x  0 , t  0 but does not affect the shape of the
y( x, t ) versus x or t graph.
 cos(kx  t   )
  sin(kx  t   ) .
t
EXECUTE: (a) The graphs for each  are sketched in Figure 15.62.
y
(b)
  Asin(kx  t   )
t
SET UP:
(c) No.    4 or   3 4 would both give A
2 . If the particle is known to be moving downward, the result
of part (b) shows that cos   0, and so   3 4.
(d) To identify  uniquely, the quadrant in which  lies must be known. In physical terms, the signs of both the
position and velocity, and the magnitude of either, are necessary to determine  (within additive multiples of 2π ).
EVALUATE: The phase   0 corresponds to y  A at x  0 , t  0 .
Figure 15.62
15.63.
IDENTIFY and SET UP: Use Eq.(15.13) to replace  , and then Eq.(15.6) to replace v.
EXECUTE:
(a) Eq.(15.25): Pav  12  F 2 A2
v  F /  says
  F / v so Pav  12

F /v

F  2 A2  12 F 2 A2 / v
  2 f so  / v  2 f / v  2 /   k and Pav  12 Fk A2 , as was to be shown.
(b) IDENTIFY: For the  dependence, use Eq.(15.25) since it involves just , not k: Pav  12  F 2 A2 .
SET UP:
EXECUTE:
Pav ,  , A all constant so
F  2 is constant, and
F112  F2 22 .
2  1 (F1 / F2 )1/ 4  1(F1 / 4F1)1/ 4  1(4)1/ 4  1 / 2
 must be changed by a factor of 1/ 2 (decreased)
IDENTIFY: For the k dependence, use the equation derived in part (a), Pav  12 Fk A2 .
SET UP: If Pav and A are constant then Fk must be constant, and F1k11  F2k22 .
Mechanical Waves
EXECUTE:
15.64.
15-21
 F   
 F   1 
2
2
k2  k1  1  1   k1  1  
 k1
 k1 / 8
  k1
F

4
F
4
16
 2  2 
 1   1 / 2 
k must be changed by a factor of 1/ 8 (decreased).
EVALUATE: Power is the transverse force times the transverse velocity. To keep Pav constant the transverse
velocity must be decreased when F is increased, and this is done by decreasing  .
IDENTIFY: The wave moves in the  x direction with speed v, so to obtain y( x, t ) replace x with x  vt in the
expression for y( x,0) .
SET UP: P( x, t ) is given by Eq.(15.21).
EXECUTE: (a) The wave pulse is sketched in Figure 15.64.
(b)
for (x  vt )   L
0
h L  x  vt / L for  L  ( x  vt )  0

 
y ( x, t )  
h
L

x

vt

 / L for 0  ( x  vt )  L

0
for (x  vt )  L
(c) From Eq.(15.21):
for ( x  vt )   L
 F (0)(0)  0

2
y ( x, t ) y ( x, t )  F (h / L)(hv / L)  Fv(h / L) for  L  ( x  vt )  0
P ( x, t )   F

2
x
t
 F (h / L)(hv / L)  Fv(h / L) for 0  (x  vt )  L

for (x  vt )  L
 F (0)(0)  0
Thus the instantaneous power is zero except for L  ( x  vt )  L, where it has the constant value Fv ( h L ) 2 .
EVALUATE:
For this pulse the transverse velocity v y is constant in magnitude and has opposite sign on either side
of the peak of the pulse.
Figure 15.64
15.65.
IDENTIFY and SET UP: The average power is given by Eq.(15.25). Rewrite this expression in terms of v and 
in place of F and  .
EXECUTE:
(a) Pav  12  F 2 A2
v  F /  so F  v / 
  2 f  2 (v /  )
Using these two expressions to replace
F and  gives Pav  2 2v3 A2 /  2 ;   (6.00  103 kg)/(8.00 m)
 2 2 P 
A   2 3av   7.07 cm
 4 v  
(b) EVALUATE:
15.66.
Pav ~ v3 so doubling v increases Pav by a factor of 8.
Pav  8(50.0 W)  400.0 W
IDENTIFY: Draw the graphs specified in part (a).
SET UP: When y( x, t ) is a maximum, the slope y / x is zero. The slope has maximum magnitude when
y( x, t )  0 .
EXECUTE: (a) The graph is sketched in Figure 15.66a.
(b) The power is a maximum where the displacement is zero, and the power is a minimum of zero when the
magnitude of the displacement is a maximum.
15-22
Chapter 15
(c) The energy flow is always in the same direction.
y
(d) In this case,
 kA sin(kx  t ) and Eq.(15.22) becomes P(x,t )   Fk A2sin 2 (kx  t ). The power is now
x
negative (energy flows in the  x-direction ), but the qualitative relations of part (b) are unchanged. The graph is
sketched in Figure 15.66b.
EVALUATE: cos and sin  are 180° out of phase, so for fixed t, maximum y corresponds to zero P and
y  0 corresponds to maximum P.
Figure 15.66
15.67.
IDENTIFY and SET UP:
v  F/ . The coefficient of linear expansion  is defined by L  L0 T . This can
F/A
to give F  Y AT for the change in tension when the temperature changes by
L/L0
T . Combine the two equations and solve for  .
be combined with Y 
EXECUTE: v1  F/ , v12  F/ and F   v12
The length and hence  stay the same but the tension decreases by F  Y A T .
v2  (F  F )/  (F  Y A T )/
v22  F/  Y A T/  v12  Y  A T/
And   m / L so A/  AL/m  V/m  1/ . (A is the cross-sectional area of the wire, V is the volume of a
v12  v22
(Y/ ) T
EVALUATE: When T increases the tension decreases and v decreases.
IDENTIFY: The time between positions 1 and 5 is equal to T/2. v  f  . The velocity of points on the string is
given by Eq.(15.9).
 60 s 
SET UP: Four flashes occur from position 1 to position 5, so the elapsed time is 4 
  0.048 s . The figure
 5000 
in the problem shows that   L  0.500 m . At point P the amplitude of the standing wave is 1.5 cm.
EXECUTE: (a) T / 2  0.048 s and T  0.096 s . f  1/ T  10.4 Hz .   0.500 m .
(b) The fundamental standing wave has nodes at each end and no nodes in between. This standing wave has one
additional node. This is the 1st overtone and 2nd harmonic.
(c) v  f   (10.4 Hz)(0.500 m)  5.20 m/s .
(d) In position 1, point P is at its maximum displacement and its speed is zero. In position 3, point P is passing
through its equilibrium position and its speed is vmax   A  2 fA  2 (10.4 Hz)(0.015 m)  0.980 m/s .
length L.) Thus v12  v22   (Y T/ ) and  
15.68.
(e) v 
15.69.
F


FL
FL (1.00 N)(0.500 m)
 18.5 g .
and m  2 
m
v
(5.20 m/s) 2
EVALUATE: The standing wave is produced by traveling waves moving in opposite directions. Each point on the
string moves in SHM, and the amplitude of this motion varies with position along the string.
IDENTIFY and SET UP: There is a node at the post and there must be a node at the clothespin. There could be
additional nodes in between. The distance between adjacent nodes is  / 2, so the distance between any two nodes
is n( / 2) for n  1, 2, 3, This must equal 45.0 cm, since there are nodes at the post and clothespin. Use this in
Eq.(15.1) to get an expression for the possible frequencies f.
EXECUTE: 45.0 cm  n( / 2),   v / f , so f  n[v /(90.0 cm)]  (0.800 Hz)n, n  1, 2, 3,
EVALUATE: Higher frequencies have smaller wavelengths, so more node-to-node segments fit between the post
and clothespin.
Mechanical Waves
15.70.
15-23
IDENTIFY: The displacement of the string at any point is y( x, t )  ( ASW sin kx)sin t. For the fundamental mode
  2 L, so at the midpoint of the string sin kx  sin(2  )( L 2)  1, and y  ASW sin t. The transverse velocity is
v y  y / t and the transverse acceleration is a y  v y / t .
SET UP: Taking derivatives gives v y 
y
  ASW cos t , with maximum value v y ,max   ASW , and
t
vy
  2 ASW sin t , with maximum value ay , max   2 ASW .
t
EXECUTE:   ay,max vy,max  (8.40 103 m s2 ) (3.80 m s)  2.21103 rad s , and then
ay 
ASW  vy,max   (3.80 m s) (2.21103 rad s)  1.72 103 m .
15.71.
15.72.
(b) v   f  (2 L)( 2 )  L   (0.386 m)(2.21  103 rad s)   272 m s .
EVALUATE: The maximum transverse velocity and acceleration will have different (smaller) values at other
points on the string.
IDENTIFY: To show this relationship is valid, take the second time derivative.


SET UP:
sin t  cos t . cost   sin t .
t
t
 2 y ( x, t )  2

 2 [( ASW sin kx)sin t ]   [( ASW sin kx)cos t ]
EXECUTE: (a)
t 2
t
t
 2 y( x, t )
  2 ( Asw sin kx)sin t    2 y ( x, t ) . This equation shows that ay   2 y . This is characteristic of
t 2
simple harmonic motion; each particle of the string moves in simple harmonic motion.
(b) Yes, the traveling wave is also a solution of this equation. When a string carries a traveling wave each point on
the string moves in simple harmonic motion.
EVALUATE: A standing wave is the superposition of two traveling waves, so it is not surprising that for both
types of waves the particles on the string move in SHM.
IDENTIFY and SET UP: Carry out the analysis specified in the problem.
EXECUTE: (a) The wave moving to the left is inverted and reflected; the reflection means that the wave moving
to the left is the same function of x, and the inversion means that the function is  f ( x).
(b) The wave that is the sum is f ( x)  f (x) (an inherently odd function), and for any f , f (0)  f (0)  0.
(c) The wave is reflected but not inverted (see the discussion in part (a) above), so the wave moving to the left in
Figure 15.21 in the textbook is  f ( x).
dy d
df ( x) df ( x) df ( x) df ( x) d ( x) df df
.
 ( f ( x)  f ( x)) 





dx dx
dx
dx
dx
d ( x) dx
dx dx x  x
At x  0 , the terms are the same and the derivative is zero.
EVALUATE: Our results verify the behavior shown in Figures 15.20 and 15.21 in the textbook.
IDENTIFY: Carry out the derivation as done in the text for Eq.(15.28). The transverse velocity is v y  y / t and
(d)
15.73.
the transverse acceleration is a y  v y / t.
(a) SET UP: For reflection from a free end of a string the reflected wave is not inverted, so
y( x, t )  y1 ( x, t )  y2 ( x, t ), where
y1 ( x, t )  A cos(kx  t ) (traveling to the left)
y2 ( x, t )  A cos(kx  t ) (traveling to the right)
Thus y( x, t )  A[cos(kx  t )  cos(kx  t )].
EXECUTE: Apply the trig identity cos(a  b)  cos a cos b sin a sin b with a  kx and b  t:
cos(kx  t )  cos kx cost  sin kx sin t and
cos(kx  t )  cos kx cost  sin kx sint.
Then y( x, t )  (2 Acos kx)cos t (the other two terms cancel)
(b) For x  0, cos kx  1 and y( x, t )  2 Acost. The amplitude of the simple harmonic motion at x  0 is 2A,
which is the maximum for this standing wave, so x  0 is an antinode.
(c) ymax  2 A from part (b).
vy 
y 
 cos t
 [(2 A cos kx)cos t ]  2 A cos kx
 2 A cos kx sin t.
t t
t
15-24
Chapter 15
At x  0, v y  2 A sin t and (v y ) max  2 A
 2 y vy
 sin t

 2 A cos kx
 2 A 2 cos kx cos t
t 2
t
t
At x  0, ay  2 A 2 cost and (ay )max  2 A 2 .
ay 
EVALUATE: The expressions for (v y ) max and ( a y ) max are the same as at the antinodes for the standing wave of a
15.74.
string fixed at both ends.
IDENTIFY: The standing wave is given by Eq.(15.28).
SET UP: At an antinode, sin kx  1 . v y ,max   A . ay ,max   2 A .
EXECUTE: (a)   v f  (192.0 m s) (240.0 Hz)  0.800 m , and the wave amplitude is ASW  0.400 cm. The
amplitude of the motion at the given points is
(i) (0.400 cm)sin ( )  0 (a node) (ii) (0.400 cm) sin(  2)  0.400 cm (an antinode)
(iii) (0.400 cm) sin(  4)  0.283 cm
(b) The time is half of the period, or 1 (2 f )  2.08  103 s.
(c) In each case, the maximum velocity is the amplitude multiplied by   2 f and the maximum acceleration is
the amplitude multiplied by  2  4 2 f 2 :
15.75.
(i) 0, 0; (ii) 6.03 m s, 9.10 103 m s 2 ; (iii) 4.27 m s, 6.43  103 m s 2 .
EVALUATE: The amplitude, maximum transverse velocity, and maximum transverse acceleration vary along the
length of the string. But the period of the simple harmonic motion of particles of the string is the same at all points
on the string.
 v 
IDENTIFY: The standing wave frequencies are given by f n  n   . v  F /  . Use the density of steel to
 2L 
calculate  for the wire.
SET UP:
EXECUTE:
For steel,   7.8  103 kg/m3 . For the first overtone standing wave, n  2 .
v
2Lf 2
 (0.550 m)(311 Hz)  171 m/s . The volume of the wire is V  ( r 2 ) L . m  V so
2
m V

  r 2  (7.8 103 kg/m3 ) (0.57 103 m)2  7.96  103 kg/m . The tension is
L
L
F   v 2  (7.96  103 kg/m)(171 m/s)2  233 N .
EVALUATE: The tension is not large enough to cause much change in length of the wire.

15.76.
IDENTIFY: The mass and breaking stress determine the length and radius of the string. f1 
v
, with v 
2L
F

.
SET UP: The tensile stress is F /  r 2 .
EXECUTE: (a) The breaking stress is F 2  7.0 108 N m2 and the maximum tension is F  900 N, so solving
r
900 N
 6.4  104 m . The mass and density are fixed,
for r gives the minimum radius r 
 (7.0  108 N m 2 )
  M2 so the minimum radius gives the maximum length
r L
L
M
4.0 103 kg

 0.40 m .
2
 r   (6.4 104 m)2 (7800 kg m3 )
F  1 F . Assuming the maximum length of the
(b) The fundamental frequency is f1  1 F  1
2L  2L M L 2 ML
string is free to vibrate, the highest fundamental frequency occurs when F  900 N and
900 N
 376 Hz .
(4.0  103 kg)(0.40 m)
EVALUATE: If the radius was any smaller the breaking stress would be exceeded. If the radius were greater, so
the stress was less than the maximum value, then the length would be less to achieve the same total mass.
IDENTIFY: At a node, y( x, t )  0 for all t. y1  y2 is a standing wave if the locations of the nodes don't depend on t.
SET UP: The string is fixed at each end so for all harmonics the ends are nodes. The second harmonic is the first
overtone and has one additional node.
f1  12
15.77.
Mechanical Waves
EXECUTE: (a) The fundamental has nodes only at the ends, x  0 and x  L.
(b) For the second harmonic, the wavelength is the length of the string, and the nodes are at
x  0, x  L 2 and x  L.
(c) The graphs are sketched in Figure 15.77.
(d) The graphs in part (c) show that the locations of the nodes and antinodes between the ends vary in time.
EVALUATE: The sum of two standing waves of different frequencies is not a standing wave.
Figure 15.77
15-25
15-26
Chapter 15
15.78.
IDENTIFY:
v
. The buoyancy force B that the water exerts on the object reduces the tension in the wire.
2L
B   fluidVsubmerged g .
SET UP:
f1 
For aluminum, a  2700 kg/m3 . For water,  w  1000 kg/m3 . Since the sculpture is completely
submerged, Vsubmerged  Vobject  V .
EXECUTE: L is constant, so
f air f w

and the fundamental frequency when the sculpture is submerged is
vair vw
v 
f w  f air  w  , with fair  250.0 Hz . v 
 vair 
F

so
vw

vair
Fw
. When the sculpture is in air, Fair  w  mg  aVg .
Fair
When the sculpture is submerged in water, Fw  w  B  ( a  w )Vg .
vw

vair
a   w
and
a
1000 kg/m3
 198 Hz .
2700 kg/m3
EVALUATE: We have neglected the buoyant force on the wire itself.
IDENTIFY: Compute the wavelength from the length of the string. Use Eq.(15.1) to calculate the wave speed and
then apply Eq.(15.13) to relate this to the tension.
(a) SET UP: The tension F is related to the wave speed by v  F /  (Eq.(15.13)), so use the information given
to calculate v.
f w  (250.0 Hz) 1 
15.79.
EXECUTE:  / 2  L
  2L  2(0.600 m)  1.20 m
Figure 15.79
v  f   (65.4 Hz)(1.20 m)  78.5 m/s
  m / L  14.4  103 kg/0.600 m  0.024 kg/m
Then F   v 2  (0.024 kg/m)(78.5 m/s) 2  148 N.
(b) SET UP: F  v2 and v  f  give F   f 2 2 .
 is a property of the string so is constant.
 is determined by the length of the string so stays constant.
 ,  constant implies F / f 2   2  constant, so F1 / f12  F2 / f 22
2
15.80.
f 
 73.4 Hz 
EXECUTE: F2  F1  2   (148 N) 
  186 N.
 65.4 Hz 
 f1 
F  F1 186 N  148 N

 0.26  26%.
The percent change in F is 2
F1
148 N
EVALUATE: The wave speed and tension we calculated are similar in magnitude to values in the Examples. Since
the frequency is proportional to F , a 26% increase in tension is required to produce a 13% increase in the
frequency.
IDENTIFY and SET UP: Consider the derivation of the speed of a longitudinal wave in Section 15.4.
EXECUTE: (a) The quantity of interest is the change in force per fractional length change. The force constant k 
is the change in force per length change, so the force change per fractional length change is k L, the applied force
at one end is F  (k L)(v y v ) and the longitudinal impulse when this force is applied for a time t is k Lt v y v . The
2
change in longitudinal momentum is ((vt ) m L)v y and equating the expressions, canceling a factor of t and solving
for v gives v 2  L2 k  m.
(b) v  (2.00 m) (1.50 N m) (0.250 kg)  4.90 m s
EVALUATE: A larger k  corresponds to a stiffer spring and for a stiffer spring the wave speed is greater.
Mechanical Waves
15.81.
15-27
IDENTIFY: Carry out the analysis specified in the problem.
SET UP: The kinetic energy of a very short segment x is K  12 (m)vy2 . v y  y / t . The work done by the
tension is F times the increase in length of the segment. Let the potential energy be zero when the segment is
unstretched.
2
2
K (1 2)mv y 1  y 
EXECUTE: (a) uk 

   .
x
m 
2  t 
1
y
(b)
  A sin(kx  t ) and so uk   2 A2 sin 2 (kx  t ).
t
2
y
(c) The piece has width x and height x , and so the length of the piece is
x
12
12
2

  y 2 
 1  y 2 
 y  
2
 (x)   x    x 1      x 1     , where the relation given in the hint has been used.
 x  

  x  
 2  x  
2
1
x 1  2 (y / x)   x 1  y 2
(d) up  F 
 F  .
x
2  x 
1
y
(e)
 kA sin(kx  t ), and so up  Fk 2 A2 sin 2 (kx  t ) .
x
2
(f ) Comparison with the result of part (c) with k 2   2 v 2   2  F shows that for a sinusoidal wave u k  u p .
(g) The graph is given in Figure 15.81. In this graph, uk and up coincide, as shown in part (f ) . At y  0, the string
is stretched the most, and is moving the fastest, so uk and up are maximized. At the extremes of y, the string is
unstretched and is not moving, so uk and up are both at their minimum of zero.
P
.
v
EVALUATE: The energy density travels with the wave, and the rate at which the energy is transported is the
product of the density per unit length and the speed.
(h) uk  up  Fk 2 A2 sin 2 (kx  t )  Fk ( v) A2sin 2 (kx  t ) 
Figure 15.81
15.82.
IDENTIFY: Apply
F
y
 0 to segments of the cable. The forces are the weight of the diver, the weight of the
segment of the cable, the tension in the cable and the buoyant force on the segment of the cable and on the diver.
SET UP: The buoyant force on an object of volume V that is completely submerged in water is B  waterVg .
EXECUTE: (a) The tension is the difference between the diver’s weight and the buoyant force,
F  (m   waterV ) g  (120 kg  (1000 kg m 3 )(0.0800 m 3 )(9.80 m s 2 ))  392 N.
(b) The increase in tension will be the weight of the cable between the diver and the point at x, minus the buoyant
force. This increase in tension is then
  x   ( Ax)  g  (1.10
kg m  (1000 kg m3 ) (1.00 102 m) 2 )(9.80 m s ) x  (7.70 N m) x . The tension as a
function of x is then F ( x)  (392 N)  (7.70 N m) x.
2
15-28
Chapter 15
(c) Denote the tension as F ( x)  F0  ax, where F0  392 N and a  7.70 N/m. Then the speed of transverse
waves as a function of x is v  dx  ( F0  ax )  and the time t needed for a wave to reach the surface is found
dt
from t   dt  
μ
dx

dx.
dx dt
F0  ax
Let the length of the cable be L, so t   
L
0
dx
2
 
F0  ax
a
F0  ax
L
0

2 
a


F0  aL  F0 .
2 1.10 kg m
( 392 N  (7.70 N m)(100 m)  392 N)  3.98 s.
7.70 N m
EVALUATE: If the weight of the cable and the buoyant force on the cable are neglected, then the tension would
F
392 N
L

 18.9 m/s and t   5.92 s . The
have the constant value calculated in part (a). Then v 

1.10 kg/m
v
weight of the cable increases the tension along the cable and the time is reduced from this value.
IDENTIFY: The tension in the rope will vary with radius r.
SET UP: The tension at a distance r from the center must supply the force to keep the mass of the rope that is
further out than r accelerating inward. The mass of this piece in m L  r , and its center of mass moves in a circle of
L
L

r
radius
.
2
2
 L  r  2  L  r  m 2 2
EXECUTE: T (r )  m
 

( L  r ). The speed of propagation as a function of distance is


L   2  2L

t
15.83.
dr
T (r )
TL 



L2  r 2 , where dr  0 has been chosen for a wave traveling from the center to
dt
dt

m
2
the edge. Separating variables and integrating, the time t is
v(r ) 
2 L dr
.
ω  0 L2  r 2
The integral may be found in a table, or in Appendix B. The integral is done explicitly by letting
dr
r
2

   arcsin , and t 
arcsin(1) 
.
r  L sin θ, dr  L cos θ dθ, L2  r 2  L cos θ, so that 
2
2
L

 2
L r
EVALUATE: An equivalent method for obtaining T (r ) is to consider the net force on a piece of the rope with
t   dt 
length dr and mass dm  dr m L . The tension must vary in such a way that
dT
 (mω2 L)rdr. This is integrated to obtained T (r )  (mω2 2 L)r 2  C , where
dr
C is a constant of integration. The tension must vanish at r  L, from which C  (mω2 L 2) and the previous result
is obtained.
IDENTIFY: Carry out the calculation specified in part (a).
y
y
 kASW cos kx sin t,
  ASW sin kx cos t . sin cos  12 sin 2 .
SET UP:
x
t
EXECUTE: The instantaneous power is
T (r )  T (r  dr )  ω2rdm, or
15.84.
1 2
2
P  FASW
k (sin kx cos kx)(sin t cos t )  FASW
k sin(2kx)sin(2t ).
4
(b) The average value of P is proportional to the average value of sin(2t ), and the average of the sine function is
zero; Pav  0.
(c) The graphs are given in Figure 15.84. The waveform is the solid line, and the power is the dashed line. At time
t  0 , y  0 and P  0 and the graphs coincide.
(d) When the standing wave is at its maximum displacement at all points, all of the energy is potential, and is
concentrated at the places where the slope is steepest (the nodes). When the standing wave has zero displacement,
all of the energy is kinetic, concentrated where the particles are moving the fastest (the antinodes). Thus, the
energy must be transferred from the nodes to the antinodes, and back again, twice in each cycle. Note that P is
greatest midway between adjacent nodes and antinodes, and that P vanishes at the nodes and antinodes.
Mechanical Waves
15-29
EVALUATE: There is energy flow back and forth between the nodes, but there is no net flow of energy along the
string.
Figure 15.84
15.85.
IDENTIFY:
SET UP:
For a string, f n  n F .
2L 
For the fundamental, n  1. Solving for F gives F   4 L2 f 2 . Note that    r 2  , so
   (0.203  103 m) 2 (7800 kg/m3 )  1.01 10 3 kg/m .
EXECUTE: (a) F  (1.01 103 kg/m)4(0.635 m) 2 (247.0 Hz) 2  99.4 N
(b) To find the fractional change in the frequency we must take the ratio of f to f : f 
1 F
and
2L 
 1
 1 F
1 
1
1
1 1 F
f   
F2 
 F2 
   

 2L 
2
L

2
L

2
L
2 F




 
1
f 2 L 

Now divide both sides by the original equation for f and cancel terms:
f
1
2L
1 F
2 F 1 F

.
2 F
F

(c) The coefficient of thermal expansion  is defined by l  l0T . Combining this with Y 
F /A
gives
l/l0
F   Y AT . F    2.00  1011 Pa 1.20  105 /C   (0.203 103 m)2 (11 C)  3.4 N . Then F/F  0.034 ,
f / f  0.017 and f  4.2 Hz . The pitch falls. This also explains the constant tuning in the string sections of
symphonic orchestras.
EVALUATE: An increase in temperature causes a decrease in tension of the string, and this lowers the frequency
of each standing wave.