Chapter 11 WAVES - University of Victoriaweb.uvic.ca/~jalexndr/192Ch11soln.pdf · Chapter 11: Waves...

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Chapter 11

WAVES

Conceptual Questions

1. Wrapping a thick coil of copper wire around a piano string increases the string’s mass density and therefore decreases the speed of waves traveling along it. The fundamental wavelength is fixed by the length of the string—the decreased wave velocity must therefore be accompanied by a decrease in the frequency at which the string vibrates.

2. Piano, guitar, and violin strings produce transverse waves when they are plucked or bowed—they do not produce longitudinal sound waves. The transverse motion of the strings causes longitudinal sound waves to be produced in the surrounding air.

3. The wavelength of the fundamental standing wave on a cello string depends only upon the length of the string. The wavelength of the sound waves produced by the vibrating cello string is determined by the frequency of the string’s vibration. This frequency is proportional to the string’s wave velocity, which depends upon both the mass per unit length and the tension of the string.

4. (a) The wavelength of the fundamental will decrease.

(b) The frequency of the fundamental will increase.

(c) The wave velocity is constant, thus the time for a pulse to travel the length of the string will decrease.

(d) The maximum velocity of a point on the string is proportional to the frequency and will therefore increase.

(e) The maximum acceleration of a point on the string is proportional to the square of the frequency and will therefore increase.

5. Words spoken by two people at the same time are comprehensible because sound waves travel through each other—interfering while superimposed, but returning to their original waveform as they again separate.

6. (a) Increasing the tension of the string changes the speed of transverse waves and thus the frequency.

(b) Pressing her finger on the string changes the string length, which changes the wavelength and hence the frequency.

(c) Each string has a different tension and mass per unit length; thus the speed of transverse waves differs from string to string. The fundamental wavelength remains the same because the strings are all of the same length, but the frequencies are different.

7. A transverse wave is produced by a disturbance that tends to shear the medium, separating layers at right angles to the direction of the wave velocity.

Chapter 11: Waves College Physics

512

8. In the figure below, the shape of a wave traveling to the right is depicted at two times separated by a short interval. Arrows indicate the direction of the string’s movement.

v

9. Since transverse waves do not travel through the core while longitudinal waves do, some part of the core is a molten, viscous liquid that cannot support the transmission of a transverse wave. A longitudinal wave can create compressions and rarefactions in the liquid and travel on through.

10. The electrical signal mimicking the noise is modified electronically to produce a wave that is out of phase with the noise. When played in the headphones it interferes destructively with the noise, thereby canceling it.

11. If the wires going to one speaker are reversed, the sound waves emitted will be out of phase with those from the other speaker. For a listener situated midway between the speakers, the sounds from the two speakers will interfere destructively, and the sound level will be noticeably reduced. If the listener moves slightly, the difference in the path length from the two speakers changes. If this change in path length is small compared to the wavelength, the waves still interfere destructively to a large extent. For this reason there is a noticeably weaker bass (long wavelength) over a large area between the speakers, while high frequencies (short wavelengths) are less affected.

Problems

1. Strategy Form a proportion with the intensities, treating the Sun as an isotropic source. Use Eq. (11-1). Solution Find the intensity of the sunlight that reaches Jupiter.

2J

2E

2 224J E E 2 2

J E2E JJ4

1, so (1400 W m ) 52 W m .

5.2

Pr

Pr

I r rI I

I rr

2. (a) Strategy It takes half the time for the sound to cross the valley as it does to make the round trip.

Solution The cliff is (343 m s)(0.75 s) 260 mx v t away.

(b) Strategy Treat the radio and echo as isotropic sources. Use Eq. (11-1) and form a proportion. Solution Find the intensity of the music arriving at the cliff.

2 22 2cliff c r r 5 2 10 2

cliff radio2 2radio cr c

(4 ) 1, so (1.0 10 W m ) 1.5 10 W m .

257.25(4 )

I P r r rI I

I rP r r

3. Strategy Form a proportion with the intensities, treating the jet airplane as an isotropic source. Use Eq. (11-1). Solution Find the intensity of the sound waves at the ears of the person.

2 2 222 2 1 1 2 2 2

2 121 2 21

(4 ) 5.0 m, so (1.0 10 W m ) 170 mW m .

120 m(4 )

I P r r rI I

I r rP r

College Physics Chapter 11: Waves

513

4. Strategy The power equals the intensity times the area. Solution Find the power radiated by the jet airplane in the form of sound waves.

22 2 25.0 m

(1.0 10 W m )4 (120 m) 31 kW120 m

P IA

5. Strategy The power equals the intensity times the area. Solution Find the rate at which the Sun emits electromagnetic waves.

2 3 2 11 2 26E(4 ) 4 (1.4 10 W m )(1.50 10 m) 4.0 10 WP IA I R

6. Strategy Refer to the figure. Use the definition of average speed.

Solution

(a) 1.80 m 1.50 m

1.5 m s0.20 sx

xv

t

(b) 8.7 cm 4.5 cm

21 cm s0.20 sy

yv

t

7. Strategy Refer to the figure. Use the definition of average speed.

Solution

(a) Find the speed. 1.80 m 1.50 m

1.5 m/s0.20 sx

xv

t

Find the position.

f i 1.80 m (1.5 m s)(3.00 s 0.20 s) 6.0 mx x v t

(b) f if i

4.00 m 1.80 m0.20 s 1.7 s

1.5 m sx

x xt t

v

8. Strategy Use Eq. (11-4). Solution Find the linear mass density of the cord.

2 2

75 N, so 3.8 g m .

(140 m s)

F Fv

v

9. Strategy Use Eq. (11-4). Solution Find the speed of the transverse waves on the string.

3

90.0 N168 m s

3.20 10 kg m

Fv


514

10. Strategy Use Eq. (11-2) to find the wave speeds for both strings. Use the speeds and the lengths of the strings to find the additional time required for the slower wave to reach the end of its string. Solution

1 11

1

(180.0 N)(15.0 m)186 m s

0.0780 kg

F Lv

m and 2 2

22

(160.0 N)(15.0 m)203 m s.

0.0580 kg

F Lv

m

The pulse moves faster on the second string. Find the time difference.

1 21 2 1 2

1 1 1 1, so (15.0 m) 6.9 ms .

186.05 m s 203.42 m s

x xx v t t t x

v v v v

11. Strategy Use Eq. (11-2) for the speed of the transverse waves. Solution The weight of the string divided by the load is

43

0.25 N2.5 10 0.025%.

1.00 10 N

The weight of the string is negligible since the result will be limited to two significant figures (by 0.25 N). Find the time it takes the wave pulse to travel to the upper end of the string.

3

(0.25 N)(10.0 m)16 ms

(1.00 10 N)(9.80 N kg)

y L mL mgLt

v F FgFL m

12. Strategy Use Eq. (11-6). Solution Find the speed of the wave.

(500.0 Hz)(0.500 m) 250 m sv f

13. Strategy Use Eq. (11-5). Solution Find the wavelength.

3(75.0 m s)(5.00 10 s) 0.375 mvT

14. Strategy Use Eq. (11-6). Solution Find the frequency.

2

120 m s400 Hz

30.0 10 m

vf

15. Strategy Use Eq. (11-6).

Solution Find the frequencies.

(a) 340 m s

340 Hz1.0 m

vf

(b) 8

83.0 10 m s3.0 10 Hz

1.0 m

vf


515

16. Strategy Use Eq. (11-6). Solution Find the range of visible electromagnetic waves.

8 814 14

1 27 71 2

3.0 10 m s 3.0 10 m s7.5 10 Hz and 4.3 10 Hz.

4.0 10 m 7.0 10 m

v vf f

The frequency range is 14 144.3 10 Hz to 7.5 10 Hz .

17. Strategy Use Eq. (11-6). Solution Find the frequency with which the buoy bobs up and down.

2.5 m s0.33 Hz

7.5 m

vf

18. Strategy and Solution If another swimmer were 9.6 m away from you, your motions would be the same (in phase). For periodic motion, the motion half a wavelength from any point along a line parallel to the motion of the waves is opposite to the motion at that point, so the other swimmer should be half a wavelength away, or 9.6 m

4.8 m .2

19. Strategy Use Eq. (11-7). Solution Compute the wave speed.

5.0 rad s0.83 cm s

6.0 rad cmv

k

20. Strategy Read the amplitude from the equation of the wave. Use Eq. (11-7) to find the wavelength.

Solution

( , ) (3.5 cm)sin (66 cm s) sin( ),3.0 cm

y x t x t A kx t

so we have

(a) 3.5 cmA

(b) 3.0 cm

2 26.0 cm

k

21. Strategy The wave on the string is of the form ( , ) sin( ).y x t A t kx Use the equation, the given information, Eq. (11-7), and the relationship between period and angular frequency to find the amplitude, wavelength, period, and wave speed.

Solution

(a) 4.0 mmA

(b) 1

2 21.0 m

6.0 mk

(c) 2 1

2 20.010 s

6.0 10 sT


516

(d) 2 1

1

6.0 10 s100 m s

6.0 mv

k

(e) Since the signs of and t kx are opposite, the wave travels in the +x-direction (to the right).

22. (a) Strategy Use Eq. (10-21). Solution Find the maximum transverse speed of a point on the string.

m (130 rad s)(0.0220 m) 2.9 m sv A

(b) Strategy Use Eq. (10-22). Solution Find the maximum transverse acceleration of a point on the string.

2 2 2m (130 rad s) (0.0220 m) 370 m sa A

(c) Strategy Use Eq. (11-7). Solution Find the wave speed.

130 rad s8.7 m s

15 rad mv

k

(d) Strategy Consider why transverse speed and wave speed are different. Solution The answer to part (c) is different from the answer to part (a) because

the motion of the particles on the string is not the same as the motion of the wave along the string .

23. Strategy The equation for a transverse sinusoidal wave moving in the negative x-direction can be written in the form ( , ) sin( ).y x t A kx t Use Eq. (11-7) to find the angular frequency and the wavenumber. Solution 0.120 m, 0.300 m, 6.40 m s, and ( , ) sin( ).A v y x t A t kx

1 12 2 (6.40 m s) 2 2134 s and 20.9 m .

0.300 m 0.300 m

vk

Thus, the equation is 1 1( , ) (0.120 m)sin[(134 s ) (20.9 m ) ] .y x t t x

24. Strategy The equation for a transverse sinusoidal wave moving in the positive x-direction and in the negative y-direction in the next instant of time according to the situation given in the problem statement can be written in the form ( , ) sin( ).y x t A kx t Use Eqs. (10-21) and (11-7). Solution Find the wave speed v by finding the maximum speed of a point on the string.

m5.00 5.00v v A

Find the wave number. 1 1

8.00 rad m5.00 5.00 5.00(0.0250 m)

kv A A

The equation for the transverse sinusoidal wave is ( , ) (2.50 cm)sin[(8.00 rad m) (2.90 rad s) ] .y x t x t


517

25. Strategy The maximum y-value is the amplitude. Find the horizontal distance for which the wave repeats to find the wavelength. The wave speed is the distance the wave travels divided by the time interval it took to travel that distance. Use Eq. (11-6) to find the frequency of the wave. The period is the reciprocal of the frequency.

Solution

(a) max 2.6 cm, so 2.6 cm .y A

(b) 16 m 2 m 14 mx

(c) 7.5 m 5.5 m

20 m s0.10 s

xv

t

(d) 20 m s

1.4 Hz14 m

vf

(e) 1 14 m

0.70 s20 m s

Tf

26. (a) Strategy The maximum y-value is the amplitude. Find the horizontal distance for which the wave repeats to find the wavelength. The wave speed is the distance the wave travels divided by the time interval it took to travel that distance. Solution Plot ( , ) (4.0 cm)sin[(378 rad s) (314 rad cm) ]y x t t x at the two given times.

0.030.01 0.02

4

0

2

y (cm)

x (cm)

t = 0 t = s1

480

2

4

max 4.0 cm, so 4.0 cm ; 0.020 cm 0 0.020 cm ;

0.0025 cm 01.2 cm s

1 480 s

y A x

xv

t

(b) Strategy Set 0;x then the function to plot is ( , ) (4.0 cm)sin[(378 rad s) ].y x t t The period of the vibration is equal to the time interval for which the wave repeats. Multiply the wave speed by the period of the vibration and compare to the wavelength found in part (a). Solution

4

0

2

2

4

y (cm)

t (s)

x = 0

π

378

π

189

s 16.6 ms and (1.2 cm s) s 0.020 cm .189 189

T vT


518

27. Strategy Use Eqs. (10-21) and (10-22). Plot the graphs. Solution Find the maximum speed and maximum acceleration of a point on the string.

2 2 2m m(4.0 rad s)(0.0050 m) 0.063 m s and (4.0 rad s) (0.0050 m) 0.79 m s .v A a A

Plot the graphs. 1(0, ) (0.0050 m)cos[(4.0 s ) ]y t t

0.25 0.50

0.0050

0

0.0050

y (m)

x = 0

t (s)

yv leads y by 14

cycle; 1(0, ) (0.063 m/s)sin[(4.0 s ) ]yv t t

0.25 0.50

0.063

0

0.063

vy (m/s)

t (s)

x = 0

ya leads yv by 14

cycle; 2 1(0, ) (0.79 m/s )cos[(4.0 s ) ]ya t t

0.25 0.50

0.79

0

0.79

ay (m/s2)

t (s)

x = 0

28. Strategy Use Eqs. (10-21) and (10-22). yv leads y by 1/4 cycle, so yv is a cosine function. Plot the graphs.

Solution 1 1( , ) (1.2 mm)sin[(2.0 s ) (0.50 m ) ];y x t t x calculate the maximum speed.

1 1m (2.0 s )(1.2 mm) 7.5 mm/s, so (0, ) (7.5 mm s)cos(2.0 s ) at 0.yv A v t t x

1.0

1.2

0

1.2

y (mm)

t (s)

x = 0

0.5

0.5 1.0

7.5

0

7.5

vy (mm/s)

t (s)

x = 0


519

29. (a) Strategy Substitute t = 0, 0.96 s, and 1.92 s into 1 1( , ) (0.80 mm)sin[( 5.0 cm ) ( 6.0 s ) ]y x t x t and graph the resulting equations. Solution The three equations are:

1( , 0) (0.80 mm)sin[( 5.0 cm ) ],y x x 1( , 0.96 s) (0.80 mm)sin[( 5.0 cm ) 0.50], andy x x

1( , 1.92 s) (0.80 mm)sin[( 5.0 cm ) 1.0].y x x Find the wavelength.

2 210 cm

(5.0 cm)k

The amplitude of the wave is A = 0.80 mm. The first graph (solid) begins at the origin. The second graph (dashed) is shifted to the right by (5.0 cm 0.50) 0.80 cm. The third graph (dotted) is shifted to the right twice as far as the second graph, or 1.6 cm.

The graphs are shown:

10

0.80

0

0.80

y (mm)

x (cm)

t = 0

5.0

t = 0.96 s

t = 1.92 s

(b) Strategy Substitute t = 0, 0.96 s, and 1.96 s into 1 1( , ) (0.50 mm)sin[( 5.0 cm ) ( 6.0 s ) ]y x t x t and graph the resulting equations. Solution The three equations are:

1( , 0) (0.50 mm)sin[( 5.0 cm ) ],y x x 1( , 0.96 s) (0.50 mm)sin[( 5.0 cm ) 0.50], andy x x

1( , 1.92 s) (0.50 mm)sin[( 5.0 cm ) 1.0].y x x Find the wavelength.

2 210 cm

(5.0 cm)k

The amplitude of the wave is A = 0.50 mm. The first graph (solid) begins at the origin. The second graph (dashed) is shifted to the left by (5.0 cm 0.50) 0.80 cm. The third graph (dotted) is shifted to the left twice as far as the second graph, or 1.6 cm. The graphs are shown:

10

0.50

0

0.50

y (mm)

x (cm)

t = 0

5.0

t = 0.96 s

t = 1.92 s

(c) Strategy Refer to the results of parts (a) and (b). Solution The graphs obtained in part (a) move to the right as time progresses, so

( , ) (0.80 mm)sin ( )y x t kx t represents a wave traveling in the +x-direction. The graphs obtained in part

(b) move to the left as time progresses, so ( , ) (0.50 mm)sin ( )y x t kx t represents a wave traveling in the

–x-direction.


520

30. (a) Strategy and Solution Since the argument of the cosine function is t kx (both terms are positive), the wave is moving to the left.

(b) Strategy The maximum y-value is the amplitude. The wave repeats every 4.0 cm, so the wavelength is 0.040 m. Use Eq. (11-7) to find the angular frequency and wavenumber. Solution

max2 2 (10.0 m s)

2.0 mm, so 2.0 mm ; 1600 rad s ;0.040 m

2 2160 rad m

0.040 m

vy A

k

(c) Strategy Choose the point (x, y) = (0, 0). 0 cos[ (0)], so 0 cos .A t k t cos 0 when 2,t t n where n is an odd integer. The smallest nonnegative n is 1 and it will give the smallest nonnegative time. Use Eq. (11-6). Solution

1 0.040 m

1.0 ms2 4 4 4 4 10.0 m s

tf f v

The period is 1 0.040 m

4.0 ms,10.0 m s

Tf v

so the three times are 1.0 ms, 5.0 ms, and 9.0 ms.

31. Strategy Compute the positions of the peaks of each pulse for the given times. Then use the principle of superposition to graph the shape of the cord for each time. Solution

t (s) Short Pulse Position Tall Pulse Position

0.15 10 cm (40 cm s)(0.15 s) 16 cm 30 cm (40 cm s)(0.15 s) 24 cm



10 20 30 40

1.5

1.0

0.5

0

y (cm)

x (cm)

t = 0.15 s

10 20 30 40

1.5

1.0

0.5

0

y (cm)

x (cm)

t = 0.25 s

10 20 30 40

1.5

1.0

0.5

0

y (cm)

x (cm)

t = 0.30 s


521

32. Strategy Compute the positions of the peaks of each pulse for the given times. Then use the principle of superposition to graph the shape of the cord for each time. Solution

t (s) Positive Pulse Position Negative Pulse Position

0.60 2.0 m (2.5 m s)(0.60 s) 3.5 m 6.0 m (2.5 m s)(0.60 s) 4.5 m

0.80 2.0 m (2.5 m s)(0.80 s) 4.0 m 6.0 m (2.5 m s)(0.80 s) 4.0 m

0.90 2.0 m (2.5 m s)(0.90 s) 4.3 m 6.0 m (2.5 m s)(0.90 s) 3.8 m

2 4 6 8

10.0

5.0

5.0

10.0

y (cm)

x (m)

t = 0.60 s

0

2 4 6 8

10.0

5.0

y (cm)

x (m)

t = 0.80 s

0

5.0

10.0

2 6 8

10.0

5.0

0

y (cm)

x (m)

t = 0.90 s

5.0

10.0

33. Strategy Sketch the sine waves. Use the principle of superposition to find the amplitudes. Let 1 sin( )y A t kx

and 2 sin( )y A t kx and use the trigonometric identity sin sin 2sin cos .2 2

Solution

(a)

300°

4

0

2

y (cm)

θ420°180°60°

2

4 Use the principle of superposition.

1 2 sin( ) sin( ) 2 sin cos sin2 2 2

y y y A t kx A t kx A t kx A t kx

where 60.0

2 cos 2(4.0 cm)cos 6.9 cm .2 2

A A


522

(b)

4

0

2

y (cm)

θ420°300°180°60°

2

4 90.0

2(4.0 cm)cos 5.7 cm2

A

34. Strategy Let 1 sin( )y A t kx and 2 sin( )y A t kx and use the trigonometric identity

sin sin 2sin cos .2 2

Use the principle of superposition.

Solution Find the traveling wave.



where 2 cos .2

A A A Find .

1 12 cos , so 2cos 120 .

2 2A A

35. Strategy Use the principle of superposition and the trigonometric identity

sin sin 2sin cos .2 2

Solution Find the traveling sine wave.



where 2 cos 6.69 cm.2

A A Find .

1 1 6.69 cmcos , so 2cos 2cos 96.0 .

2 2 2 2(5.00 cm)

A A

A A

36. Strategy f v and the frequency is the same in both mediums. Solution Find the wavelength of the light in water.

a w w 6w a

a w a, so 0.750(0.500 10 m) 375 nm .

v v v

v

37. Strategy Refer to the figure. Use xx v t and the principle of superposition.

Solution The pulse moves 1.80 m 1.50 m = 0.30 m in 0.20 s. So, the speed of the wave is

0.30 m1.5 m s.

0.20 sv When the pulse reaches the right endpoint, it is reflected and inverted. When exactly half

of the pulse has been reflected and inverted, the superposition of the incident and reflected waves results in the

cancellation of the waves 1 2( 0).y y Thus, the string looks flat at 4.0 m 1.5 m

1.7 s .1.5 m s

xt

v


523

38. Strategy Refer to the figure. Use .xx v t

Solution The pulse moves 1.80 m 1.50 m = 0.30 m in 0.20 s. So, the speed of the wave is

0.30 m1.5 m s.

0.20 sv The pulse travels to the right until it reaches the endpoint; it is then reflected and inverted.

It then travels to the left until it hits the left endpoint; it is again reflected and inverted. The pulse travels to the right until it reaches x = 1.5 m. The total distance traveled is 2.5 m + 4.0 m + 1.5 m = 8.0 m. The elapsed time is

8.0 m5.3 s .

1.5 m s

xt

v

39. Strategy The waves are coherent. Use the principle of superposition.

Solution

(a) The resulting wave will have its largest amplitude if the waves interfere constructively. The phase difference

is 0 , and the amplitude is 1 2 5.0 cm 3.0 cm 8.0 cm .A A

(b) The resulting wave will have its smallest amplitude if the waves interfere destructively. The phase difference

is 180 , and the amplitude is 1 2 5.0 cm 3.0 cm 2.0 cm .A A

(c) 8.0 cm : 2.0 cm 4 :1

40. Strategy The waves are coherent. Use the principle of superposition. Intensity is proportional to the square of the amplitude.

Solution

(a) The resulting wave will have its highest intensity when the waves interfere constructively. The phase

difference is 0 , and the amplitude is 1 2 6.0 cm 3.0 cm 9.0 cm .A A A

(b) The resulting wave will have its lowest intensity when the waves interfere destructively. The phase difference

is 180 , and the amplitude is 1 2 6.0 cm 3.0 cm 3.0 cm .A A A

(c) Form a proportion. 2 2

1 1

2 2

9.0 cm9.0

3.0 cm

I A

I A

The ratio is 9 :1 .


524

41. Strategy Intensity is proportional to the amplitude squared. For constructive interference, the amplitude of the superposition is the sum of the original amplitudes. Solution Find 1 2/ .A A

1 1

2 2

25 5.0

15 3.0

A I

A I

Find the amplitude of the superposition.

1 2 2 2 2

2 22 2

22 2

5.0 5.01

3.0 3.0

Find the intensity of the superposition.

5.0 5.0 5.01 , so 1 1 (15 mW m ) 79 mW m .

3.0 3.0 3.0

A A A A A A

I AI I

I A

42. Strategy Intensity is proportional to the amplitude squared. For destructive interference, the amplitude of the superposition is the absolute value of the difference of the original amplitudes. Solution Find 1 2/ .A A

1 1

2 2

25

28

A I

A I


1 2 2

2 23 2 2

22 2

251

28

Find the intensity of the superposition.

25 25 251 , so 1 1 (28 10 W m ) 80 W m .

28 28 28

A A A A

I AI I

I A

43. Strategy Intensity is proportional to the amplitude squared. For constructive interference, the amplitude of the superposition is the sum of the original amplitudes. For destructive interference, the amplitude of the superposition is the absolute value of the difference of the original amplitudes. For incoherent waves, the intensities add.

Solution

(a) Find 1 2/ .A A

1 1

2 2

0.040 2.0

0.090 3.0

A I

A I


1 2 2 2 2

2 22 2

22 2

2.0 2.01

3.0 3.0Find the intensity of the superposition.

2.0 2.0 2.01 , so 1 1 (0.090 W m ) 0.25 W m .

3.0 3.0 3.0

A A A A A A

I AI I

I A


525

(b) Find the amplitude of the superposition.

1 2 2

2 22 2

2

2.01

3.0Find the intensity of the superposition.

2.0 2.01 1 (0.090 W m ) 0.010 W m

3.0 3.0

A A A A

I I

(c) Find the intensity of the superposition. 2 2 2

1 2 0.040 W m 0.090 W m 0.130 W mI I I

44. Strategy The intensity minimums imply that the distance 37.1 m 25.8 m 11.3 m is equal to a whole number of wavelengths m plus one-half wavelength. Determine the number (or numbers) of wavelengths m that gives a frequency (or frequencies) between 100 Hz and 150 Hz. Use Eq. (11-6). Solution The number of wavelengths m is related to the distance between intensity minimums by

111.3 m.

2m

In terms of frequency, we have 1 343 m s

(30.35 Hz) 15.18 Hz.2 11.3 m

vf m m

Substitute 100 Hz and 150 Hz for f and solve for m to find the range of possible values. (30.35 Hz) 15.18 Hz 100 Hz, so 2.8 and (30.35 Hz) 15.18 Hz 150 Hz, so 4.4.m m m m The two possible values of m are 3 and 4. Try them both.

1 343 m s 1 343 m s(3) 3 106 Hz and (4) 4 137 Hz.

2 11.3 m 2 11.3 mf f

Both values are within the range of allowed frequencies, so the possible frequencies of the sound waves coming from the speakers are 106 Hz and 137 Hz.

45. Strategy Use Eqs. (11-2) and (11-13). Solution

1 and .2

v TLf v

L m

21 1 2 2

11 1

1, so .

2 4

T L T f Tf

L m Lm f T

Calculate the percentage reduction in the tension.

2

1

22 21 1

1 2 2

1 1 1

1 0.040100% 100% 1 100% 1 100% 7.8%

1

ff

T TT T f

T T f

46. Strategy According to Eq. (11-13), the fundamental frequency of a guitar string is directly proportional to the speed of a wave on the string. According to Eq. (11-2), the speed of a wave on the string in directly proportional to the square root of the tension in the string.

Solution f v and ,v T so ,f T where T is the tension. Therefore, since 1.15 1.07,

the frequency increases by 7% .

47. Strategy Nodes are separated by a distance of /2. Use Eq. (11-7). Solution Find the distance between adjacent nodes.

2

2distance 0.016 m

2 2 2.0 10 rad mk


526

48. Strategy The frequencies are given by (2 ).nf nv L The speed of the transverse waves is related to the tension

by .v T

Solution

(a) Find the speed of the transverse waves.

1 1, so 2 2(1.50 m)(450.0 Hz) 1350 m s .2

vf v Lf

L

(b) Find the tension.

2 2 6 2 21 1

1, so 4 4(25.0 10 kg m)(1.50 m) (450.0 Hz) 45.6 N .

2 2

v Tf T L f

L L

(c) The frequencies are the same for both mediums, but the wavelength depends upon the wave speed.

1340 m s

450.0 Hz, so 0.76 m.450.0 Hz

vf f

f

The wavelength and frequency are 0.76 m and 450.0 Hz , respectively.

49. Strategy The frequencies are given by (2 ).nf nv L The speed of the transverse waves is related to the tension

by .v T

Solution

(a) Find the frequency of the fundamental oscillation.

1 31 1 12 N

33 Hz2 2 2(1.5 m) 1.2 10 kg m

v Tf

L L

(b) Find the tension. 2 2 3 2 3 2

33

43 3 4(1.2 10 kg m)(1.5 m) (0.50 10 Hz), so 300 N .

2 2 9 9

L fv Tf T

L L

50. Strategy Use Newton’s second law and Eqs. (11-4) and (11-13).

Solution Find the tension in the string. 0, so .yF T mg T mg

Find the speed of waves on the string.

T mgv

Find the fundamental frequency.

T

mg

2.00 m

T

2

1 6

1 1 (2.20 kg)(9.80 m s )616 Hz

2 2 2(2.00 m) 3.55 10 kg m

v mgf

L L


527

51. Strategy The frequencies are given by (2 ).nf nv L The speed of the transverse waves is related to the mass per

unit length by .v T

Solution Find the mass per unit length of the guitar string.

41 2 2 2 2

1

1 82 N, so 4.5 10 kg m .

2 2 4 4(0.65 m) (329.63 Hz)

v T Tf

L L L f

52. Strategy The fundamental frequency is given by (2 )nf nv L with n = 1. Use the definition of average speed. Solution Find the fundamental frequency.

2.0 m0.050 s

1 10 Hz2 2(2.0 m)

vf

L

53. (a) Strategy and Solution All frequencies higher than the fundamental are integral multiples of the fundamental. Since there are no other frequencies between the two given, the fundamental is the difference

between those two. Thus, the fundamental frequency is 1040 Hz 780 Hz 260 Hz .

(b) Strategy Use Eqs. (11-2) and (11-13). Solution Find the total mass of the string.

1 2 21

1 1200 N, so 2.8 g .

2 2 4 4 4(260 Hz) (1.6 m)

v FL F Ff m

L L m mL f L

54. Strategy The weight of the weight equals the tension in the string. Use Eqs. (11-4) and (11-13) to find the required weight of the weight. Solution Find the weight F.

2

2

2 2 3 2 22

2 2

, so2 2 4

4 4(0.120 10 kg m)(0.42 m) (110 Hz)1.0 N, where 1, 2, 3, ... .

n

n

nv n F n Ff

L L L

L fF n n

n n

55. Strategy Use Eqs. (11-2) and (11-13). Solution

2nnv

fL

and .TL

vm

Find the total mass of the wire.

1 2 21

1 300.0 N, so 0.050 kg .

2 2 4 4 4(2.0 m)(27.5 Hz)

v TL T Tf m

L L m Lm Lf


528

56. (a) Strategy Replace each quantity with its SI units.

Solution Show that FL m has units of speed.

FL

m has units

2N m kg m s mm s,

kg kg

which are the units of speed.

(b) Strategy Set [ ] [ ] [ ]a b cF L m equal to m s , the units of speed, and determine a, b, and c.

Solution Show that no combination of L, m, and F other than FL m has units of speed.

1

2

2 0 1 1

[ ] [ ] [ ] m s

N m kg

kg mm kg

s

kg m s kg m s

a b c

a b c

a ab c

a

a c a b a

F L m

Equate exponents.

a + c = 0, so a = c. a + b = 1, so b = 1 a = 1 + c. 2a = 1, so 1 1

, ,2 2

a c and 1

.2

b

The only combination of F, L, and m that gives units of speed is 1/ 2 1/ 2 1/ 2 / ;F L m FL m therefore, the

speed of transverse waves on the string can only be /FL m times some dimensionless constant.

57. Strategy The wave speed for the 1.0-Hz waves is twice that for the 2.0-Hz waves, so it takes the 1.0-Hz waves 120 s to reach you. (120 s + 120 s = 240 s is the time it takes the 2.0-Hz waves to reach you; twice as long.) Solution Compute the distance to the boat.

(1.56 m s)(120 s) 190 mx v t

58. Strategy The wave is harmonic; ( , ) (1.2 cm)sin[(0.50 rad/s) (1.00 rad/m) ].y x t t x Use Eqs. (10-21) and (10-22). Plot the graphs. Solution Find the maximum velocity and maximum acceleration for a point on the string.

2 2 2m m(0.50 rad s)(1.2 cm) 1.9 cm s and (0.50 rad s) (1.2 cm) 3.0 cm s .v A a A

yv leads y by a quarter cycle, so yv is a cosine function.

(1.9 cm s)cos[(0.50 rad s) (1.00 rad m) ]yv t x

ya leads yv by a quarter cycle, so ya leads y by a half cycle; ya is a negative sine function.

2(3.0 cm s )sin[(0.50 rad s) (1.00 rad m) ]ya t x

The period is 2 2

4.0 s.0.50 rad s

T

1.2

0 3 6

1.2

y (cm)

t (s)

x = 0

1.9

0 3 6

1.9

x = 0

vy (cm/s)

t (s)

3.0

0 3 6

3.0

x = 0ay (cm/s2)

t (s)


529

59. Strategy Use Eq. (11-6). Solution Find the wavelength of the radio waves.

8

63.0 10 m s

3.3 m90 10 Hz

v

f

60. (a) Strategy In a longitudinal wave, the motion of particles in the medium is along the same line as the direction of propagation of the wave.

Solution

A given particle will oscillate sinusoidally in the -direction about its equilibrium position with anamplitude of 5.0 cm and a period of 1/8 s.

y

[ 10 cm (80 cm s ) 1 8 s].T v

(b) Strategy In a transverse wave, the motion of particles in the medium is perpendicular to the direction of propagation of the wave.

Solution If the wave under consideration were transverse,

the particles would oscillate along a direction perpendicular to the -axis .y

61. Strategy Speed is inversely proportional to the time of travel. Form a proportion and use .x v t Solution Relate the speeds to the times of travel.

SP

S P

10.0 km s 5.00

8.0 km s 4.0

tv

v t

Find the time for the S wave to travel from the source to the detector.

S P S S S S4.0 2.0 s

(1 0.80), so 10 s.5.00 0.20

t t t t t t t

Calculate the distance between the source and the detector.

S S (8.0 km s)(10 s) 80 kmd v t

62. Strategy Refer to the figure to determine the frequency; count the number of cycles during the indicated time period. Use Eq. (11-6) and the definition of average speed. Solution Estimate the wavelength of the seismic waves.

180,000 m30.0 s

4.0 cycles2.6 s 1.0 s

2.4 kmv

f


530

63. Strategy According to Eq. (11-13), the frequency of the string is inversely proportional to the length of the string. Solution Form a proportion.

2 1

1 2

1, so .

L ff

L L f

Relate the distance between frets to the frequencies and the length of the string.

1 11 2 1 1 1

2 21

f fL L L L L L

f f

First fret: 1

(64.8 cm) 1 3.64 cm1.0595

L

Second fret: 1

3.64 cm (64.8 cm 3.64 cm) 1 7.07 cm1.0595

Third fret: 1

7.074 cm (64.8 cm 7.074 cm) 1 10.32 cm1.0595

64. Strategy Use Eq. (11-13).

Solution Find the lowest standing wave frequencies in each situation.

(a) Since 2nnv

fL

and 1 ,2

vf

L 1 (300.0 Hz).nf nf n

n 1nf

2 600.0 Hz

3 900.0 Hz

4 1.200 kHz

(b) The lowest frequency is now 2,f and only even harmonics are allowed (always a node at the center).

n 1nf

2 600.0 Hz

4 1.200 kHz

6 1.800 kHz

8 2.400 kHz

(c) The effective length of the string is now half of the original length.

, so 2 .2 2( 2)n n nnv nv nv

f f fL L L

n 2n nf f

1 600.0 Hz

2 1.200 kHz

3 1.800 kHz

4 2.400 kHz


531

65. Strategy The fundamental frequency depends on the wavelength of the fundamental and the wave speed. The wavelength is determined by the length of the wire. The wave speed depends on the tension and the linear mass density. So we need to find out how those three quantities (length, tension, linear mass density) change when the wire is cut in half and then find the new fundamental frequency. The only numerical value we know is the original frequency, so we work by proportions. According to Eq. (11-4), the speed of waves on a wire is directly proportional to the square root of the tension. According to Eq. (11-13), the frequency of the waves on a wire is directly proportional to the speed of the waves. Therefore, the frequency is directly proportional to the square root of the tension in a wire. Use Newton’s second law.

Solution For a wire fixed at both ends, the wavelength of the

fundamental is 2L. When the wire is cut in half, then,

the wavelength is cut in half: f i1

2 . The linear

mass density does not change (half the length and half the mass), but the tension does (two wires supporting the sign instead of one).

T1

mg

T2

mg

T2

For the single wire supporting the sign (see FBD): i i0, so .yF T mg T mg

For the two wires: if f

1 12 0, so .

2 2yF T mg T mg T

The frequency is f = v/λ and the wave speed is v = /F , so 1 F

f

. Form a proportion:

f i f if 1

i f 1 f

12 1 2, so 2 2 660 Hz = 930 Hz

2

f Tf f

f T

66. (a) Strategy For a harmonic wave moving to the left (x-direction), 2 2

sin( ) sin .y A t kx A t xT

Solution Write an equation for the surface seismic waves.

32 2

( , ) (0.020 m)sin , so ( , ) (0.020 m)sin[(1.6 rad/s) (0.0016 rad/m) ] .4.0 s 4.0 10 m

y x t t x y x t t x

(b) Strategy Use Eq. (10-21). Solution Find the maximum speed of the ground as the waves move by.

m2 2 (0.020 m)

0.031 m s4.0 s

Av A

T

(c) Strategy Use Eq. (11-6). Solution Find the wave speed.

4.0 km1.0 km s

4.0 sv f

T

67. (a) Strategy Use Hooke’s law. Solution Explain why the tension in the spring is approximately proportional to the length.

0 0Hooke’s law: ( ) for .T k x x kx x x


532

(b) Strategy Use Eq. (11-2) and the result of part (a). Solution Find the time it takes the wave to travel the length of the spring.

2 2 since , and , so or .

TL kL L kL mv T kx kL L v t t

m m t m k

Since neither m nor k change, the increase in length does not affect the time of travel, so 4.00 s .t

68. Strategy Use dimensional analysis.

Solution has units m. g has units 2m s . g has units 2 2m s . g has units m s. So, .v g

69. Strategy Use dimensional analysis.

Solution has units 2N m kg s . has units 3kg m . has units m. ( ) has units

2 2

3 2

kg s m.

m kg m s

1/ 2[ ( )] has units m s. So, .v

Since v depends upon , surface waves are

dispersive.

70. Strategy Use the properties of traveling waves to answer the questions concerning the seismic wave described by the equation ( , ) (7.00 cm)cos[(6.00 rad cm) (20.0 rad s) ].y x t x t

Solution

(a) Because and kx t have the same sign, the wave is moving to the left .

(b) The particles in the medium move a distance from their equilibrium positions equal to the amplitude of the

wave. Thus, they move 7.00 cm .

(c) The frequency of this wave is 120.0 s

10.0 Hz .2 2

f

(d) The wavelength of this wave is 1

2 20.333 cm .

6.00 cmk

(e) The wave speed is 1

1

20.0 s3.33 cm s .

6.00 cmv

k

(f) A particle that is at 7.00 cm and 0 when 0y x t

oscillates sinusoidally along the -axis about = 0 with an amplitude of 7.00 cm .y y

(g) Since the motion of particles in the medium is perpendicular to the direction of propagation of the wave, the

wave is a transverse wave.

71. Strategy and Solution Refer to the figure.

(a) Since the wave is moving to the left, the peak of the pulse has not yet reached the position of point A. So, point A is moving upward.


533

(b) The peak has passed point B, so point B is moving downward.

(c) The slope of the string is larger at point A than at point B, so the speed of the string segment is larger at point A.

72. Strategy The position of the particle will follow the shape of the wave. The velocity is a bit more complicated. As the wave passes the point under consideration, the particle moves upward rapidly until it reaches the “top” of the wave. When it reaches the top, its velocity is instantaneously zero. The point then moves downward at an average velocity less than that with which it rose. The velocity is equal to the slope of the position graph for any time t. Solution The plot of the position as a function of time:

t

x

The plot of the velocity as a function of time:

t

vx

73. Strategy Destructive interference occurs when the path length difference of the two sound waves is an odd multiple of half of the wavelength.

Solution The wavelength is / (340 m/s)/(680 Hz) 0.50 m.v f The largest possible path length difference

is equal to the distance between the speakers, 1.5 m. / 2 0.25 m, so the path length differences that cause destructive interference are 0.25 m, 0.75 m, and 1.25 m. Let the speakers lie along the x-axis at 0.75 m.x Then the path length difference is zero along the y-axis and 1.5 m along the x-axis. As the listener walks along the circle of radius 1 m, the path length difference varies from 0 to 1.5 m. The path length difference equals 0.25 m, 0.75 m, and 1.25 m once for each quadrant of the circle (three occurrences of destructive interference). There are four quadrants, so the listener observes destructive interference at 12 points along the circle.

74. (a) Strategy Substitute t = 0, 1.0 s, and 2.0 s into

1 1( , ) (5.0 cm) sin[( 5.0 cm ) ( 6.0 rad s) ] sin[( 5.0 cm ) ( 6.0 rad s) ]y x t x t x t

and graph the resulting equations. Use Eq. (11-7) to find the wavelength. Solution The three equations are:

1 1 1( , 0) (5.0 cm) sin[( 5.0 cm ) ] sin[( 5.0 cm ) ] (10 cm)sin[( 5.0 cm ) ],y x x x x

1 1( , 1.0 s) (5.0 cm) sin[( 5.0 cm ) 0.52] sin[( 5.0 cm ) 0.52] , andy x x x

1 1( , 2.0 s) (5.0 cm) sin[( 5.0 cm ) 1.05] sin[( 5.0 cm ) 1.05] .y x x x

Find the wavelength. 2 2

10 cm(5.0 cm)k

The amplitude of the first (t = 0) wave is A = 10 cm. The second wave (t = 1.0 s) has a smaller amplitude than


534

the first, due to the opposite phase shifts. The third wave is even smaller in amplitude. The graphs are shown.

10

10

0

10

y (cm)

x (cm)

t = 0

5.0

t = 1.0 s

t = 2.0 s

(b) Strategy and Solution The wave doesn’t “travel” anywhere—it just oscillates up and down—so this is a standing wave.

75. Strategy Refer to Problem 78. The function is [sin( ) sin( )].y A kx t kx t Use the trigonometric identity

sin sin 2sin[( ) 2]cos[( ) 2] and the principle of superposition. Solution Use the identity.

[sin( ) sin( )] sin( ) sin( )( )

2 sin cos 2 sin( )cos( ) [2 cos( )]sin( ) sin( )2 2

y A kx t kx t A t kx A t kxkx t kx t kx t kx t

A A kx t A t kx A kx

Therefore, 2 cos( ).A A t Prove this holds for each of the amplitudes of the graphs from Problem 79.

1 1

1

1

( , ) (5.0 cm) sin[( 5.0 cm ) ( 6.0 rad s) ] sin[( 5.0 cm ) ( 6.0 rad s) ]

2(5.0 cm)sin[( 5.0 cm ) ]cos[( 6.0 rad s) ]

(10 cm)sin[( 5.0 cm ) ]cos[( 6.0 rad s) ]

y x t x t x t

x t

x t

So the maximum amplitude is 10 cm for x = 2.5 cm. At t = 1.0 s,

1(2.5 cm, 1.0 s) (10 cm)sin[( 5.0 cm )(2.5 cm)]cos[( 6.0 rad s)(1.0 s)] (10 cm)(1)cos( 6.0) 8.7 cm.y At t = 2.0 s,

1(2.5 cm, 2.0 s) (10 cm)sin[( 5.0 cm )(2.5 cm)]cos[( 6.0 rad s)(2.0 s)] (10 cm)(1)cos( 3.0) 5.0 cm.y The values correspond nicely with those shown in the graphs in the solution for Problem 79. Using the original function, we have

1

1(2.5 cm, 1.0 s) (5.0 cm){sin[( 5.0 cm )(2.5 cm) ( 6.0 rad s)(1.0 s)]

sin[( 5.0 cm )(2.5 cm) ( 6.0 rad s)(1.0 s)]} 8.7 cm

y

and 1

1(2.5 cm, 2.0 s) (5.0 cm){sin[( 5.0 cm )(2.5 cm) ( 6.0 rad s)(2.0 s)]

sin[( 5.0 cm )(2.5 cm) ( 6.0 rad s)(2.0 s)]} 5.0 cm.

y

Therefore, the amplitudes of the graphs of Problem 79 satisfy the equation 2 cos( ),A A t where A is the amplitude of the wave plotted and A is 5.0 cm.


535

76. Strategy Use the principle of superposition. Solution 1.80 m 1.50 m 0.30 mx in 0.20 s,t so 0.30 m (0.20 s) 1.5 m s.v Find the position of the peak at t = 2.2 s.

peak 1.5 m (1.5 m s)(2.2 s) 4.8 m (3.2 m;x 4.8 m 4.0 m = 0.8 m to the left)

The peak of the pulse is now inverted due to reflection.

10

0

5

5

10

y (cm)

x = 4.0 m

x = 3.2 m

x = 0

77. Strategy Use the principle of superposition. Solution 1.80 m 1.50 m 0.30 mx in 0.20 s,t so 0.30 m (0.20 s) 1.5 m s.v Find the position of the peak at t = 1.6 s.

peak i 1.5 m (1.5 m s)(1.6 s) 3.9 mx x vt

The peak of the pulse is nearly to the end of the string. The reflected pulse is below the string, so most of the height of the original pulse is cancelled.

10

0

5

y (cm)

x = 3.9 m

x = 4.0 mx = 0

Chapter 11 WAVES - University of Victoriaweb.uvic.ca/~jalexndr/192Ch11soln.pdf · Chapter 11: Waves...

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Transcript of Chapter 11 WAVES - University of Victoriaweb.uvic.ca/~jalexndr/192Ch11soln.pdf · Chapter 11: Waves...