Tài liệu Physics exercises_solution: Chapter 16 ppt

The overall effect is that for such a large bulk modulus, large pressure amplitudes are needed to produce a given displacement... b Repeating the above with the path difference an odd mu

16.1: a) λv f (344m s) (100Hz)0.344m b) if p1000p0,then A

0

1000A Therefore, the amplitude is 1.210 5m c)Since pmax BkA,increasing pmaxwhile keeping A constant requires decreasing k, and increasing π, by the same factor Therefore the new wavelength is (0.688m)(20) 6.9m, 3446.9mms 50Hz

Hz) 1000 ( Pa) 9 10 2 2 ( 2

) m 1480 ( Pa) 2 10 0 3 ( 2

v

increases both the needed pressure amplitude and the speed, but the speed is proportional

to the square root of the bulk modulus The overall effect is that for such a large bulk modulus, large pressure amplitudes are needed to produce a given displacement

16.3: From Eq (16.5), pmax BkA2π BA λ2πBA f v

Pa

108.1m)102.1(

2π  8   3 This is (8.110 3Pa) (3.010 2Pa)0.27

times smaller than the pressure amplitude at sea level (Example 16-1), so pressure amplitude decreases with altitude for constant frequency and displacement amplitude

16.5: a) Using Equation (16.7),  2  2  2

s)400m)(

8(so,)λ

ρ v B

Pa

1033.1)



v ρ L t ρ Y

Pa

1047.9)m

kg

6400



16.6: a) The time for the wave to travel to Caracas was 9min 39 s579 s and the speed was 1.085104 m s (keeping an extra figure) Similarly, the time for the wave to travel to Kevo was 680 s for a speed of 1.278104 m s, and the time to travel to Vienna was 767 s for a speed of 1.258104 m s The average speed for these three

measurements is 1.21104 m s Due to variations in density, or reflections (a subject addressed in later chapters), not all waves travel in straight lines with constant speeds b) From Eq (16.7), Bv2, and using the given value of  3.3103 kg m3 and the speeds found in part (a), the values for the bulk modulus are, respectively,

Pa

102.5and Pa105.4

16.7: Use vwater 1482m s at 20 as given in Table C,  16.1 The sound wave travels

in water for the same time as the wave travels a distance 22.0m1.20m20.8m in air, and so the depth of the diver is

sm344

sm1482m8.20m

8.20

air

v v

This is the depth of the diver; the distance from the horn is 90.8m

16.8: a), b), c) Using Eq 16.10,

molkg1002.2

K15.300KmolJ3145.841

3 2

K15.300KmolJ3145.867

3 e

K15.300KmolJ3145.867.1

d) Repeating the calculation of Example 16.5 at T 300.15K gives vair 348m s, and

so vH 3.80vair,vHe 2.94vair and vAr 0.928vair

16.9: Solving Eq (16.10) for the temperature,

hrkm6.3

sm10.85

hkm850mol)kg108.28(

2 3

p Since T T oy,m,

C106.K,

191

for T     2 and T0 273K, y13,667m(44,840ft.) Although a very high altitude for commercial aircraft, some military aircraft fly this high This result assumes a uniform decrease in temperature that is solely due to the increasing altitude Then, if we use this altitude, the pressure can be found:

,K

273

m)(13,667)

mC106(

1

) s m mol)(9.8 kg 10 8 28 ( 2

o

2 2 3

kg108.28

K)K)(294.15mol

J3145.8)(

04.1(

The same calculation with T 283.15gives344.22m s,so the increase is 0.58m s

16.11: Table 16.1 suggests that the speed of longitudinal waves in brass is much higher

than in air, and so the sound that travels through the metal arrives first The time

difference is

s

208.0)mkg(8600Pa)

1090.0(

m0.80s

m344

m0.80

3 11

Brass air

L t

mol)kg108.28(

K)K)(260.15mol

J3145.8)(

40.1(mol)

kg108.28(

K)K)(300.15mol

J3145.8)(

(The result is known to only two figures, being the difference of quantities known to three figures.)

16.13: The mass per unit length  is related to the density (assumed uniform) and the cross-section area Aby μ Aρ,so combining Eq (15.13) and Eq (16.8) with the given relations between the speeds,







900so

Αρ

F ρ

Υ

Hz220

)mkg10(8.9Pa)100.11

b) Solving for the amplitude A (as opposed to the area aπr2)in terms of the average power Pav Ιa,

2

av )2(

ρΥω

a P

A

))Hz220(2(Pa)100.11)(

mkg109.8(

)m)10(0.800(

W))1050.6(

2 10

3 3

2 -2 6

c) ω2π f Α2π(220Hz)(3.28910 8m)4.5510 5 m s

16.15: a) See Exercise 16.14 The amplitude is

mkg1000(

)mW1000.3(

2 9

3

2 6

3400

)mkg(1000Pa)

1018.2

b) Repeating the above with B  pγ 1.40105Pa and the density of air gives

m

0.100and

m10

66

A c) The amplitude is larger in air, by a factor of about

60 For a given frequency, the much less dense air molecules must have a larger

amplitude to transfer the same amount of energy

max 2

BkA p

b) From Eq (16.14),

.mW104.58s))

m344)(

mkg2.1(2(Pa)95.1(





16.18: (a) The sound level is

dB

57

or ,log

dB)(10so

,log

b) First find v, the speed of sound at 20.0C, from Table 16.1, v344m s

The density of air at that temperature is 1.20kg m3 Using Equation (16.14),

.mW1073.2

or ,s)m344)(

mkg20.1

(

2

)mN150.0(2

2 5

3

2 2 2

mW102.73logdB)10

m kg 20 1 ( 2

Pa) 10 0 6 ( 3 2 5

16.20: a) 10log 4I I 6.0dB b) The number must be multiplied by four, for an increase of 12 kids

16.21: Mom is five times further away than Dad, and so the intensity she hears is

2

251 5 of the intensity that he hears, and the difference in sound intensity levels is

dB

14log(25)

16.22:

dB25dB90dB75level)

I I

0 i

log10level)



3 5

16.24: Open Pipe:

Hz594

2

1 1

v f

Hz2972

Hz594

Hz5944

v L L

16.25: a) Refer to Fig (16.18) i) The fundamental has a displacement node at

0

and34L  and the second overtone mode has displacement nodes at

m000.1andm600.0m,

720

0

5

3L  L Displacement nodes at Fundamental: 0 First

overtone:0,2L 30.800m Second overtone:0, 2L 50.480 m, 4L 50.960m

16.26: a) 1  2  2((3440.450mm)s) 382Hz,

L v

always exact multiples of the fundamental, due to rounding

20f  so 103 rd highest harmonic heard

16.27: 4(0.17( 344 m/s)m) 506Hz, 2 3 1 1517Hz, 3 5 1 2529Hz

f

16.28: a) The fundamental frequency is proportional to the square root of the ratio M

(see Eq (16.10)), so

Hz,76700.4

8.28)57(

3)5(Hz)262(

He

air air

He air

4L and so the frequency is f1 344m s (0.56m)0.614kHz b) The length

of the column is half of the original length, and so the frequency of the fundamental mode is twice the result of part (a), or 1.23 kHz

16.30: For a string fixed at both ends, Equation 15.33, 2L nv,

n

f  is useful It is

important to remember the second overtone is the third harmonic Solving for v, 2 ,

n L

f n

vand inserting the data, v  2 635 m3588 /s, and v249m s

16.31: a) For constructive interference, the path difference d 2.00mmust be equal to

an integer multiple of the wavelength, so λn d n,

172Hz.m

2.00

sm344

v n d

vn v

Therefore, the lowest frequency is 172 Hz

b) Repeating the above with the path difference an odd multiple of half a

wavelength, f n  n 12 172Hz.Therefore, the lowest frequency is 86Hzn0

16.32: The difference in path length is xLxxL2x,or xLx 2.For destructive interference, x  n( (1 2))λ,and for constructive interference, xnλ.The wavelength is λv f 344m s (206Hz)1.670m (keeping an extra figure), and so

to have 0xL,4n3 for destructive interference and 4n4for constructive interference Note that neither speaker is at a point of constructive or destructive

interference

a) The points of destructive interference would be at x0.58m,1.42m

b) Constructive interference would be at the points x0.17m, 1.00m, 1.83m.c) The positions are very sensitive to frequency, the amplitudes of the waves will not

be the same (except possibly at the middle), and exact cancellation at any frequency is not likely Also, treating the speakers as point sources is a poor approximation for these dimensions, and sound reaches these points after reflecting from the walls , ceiling, and floor

16.33: λv f 344m s 688Hz0.500m

To move from constructive interference to destructive interference, the path

difference must change by 2.λ If you move a distance x toward speaker B, the distance

to B gets shorter by xand the difference to A gets longer by xso the path difference

changes by 2x.

2xλ 2and xλ 40.125m

16.34: We are to assume v344m s,soλv f 344m/s 172Hz2.00m If

m

r A 8.00 and r are the distances of the person from each speaker, the condition for B

destructive interference is r Br A n21λ, where n is any integer Requiring

16.35: v f 344m s 860Hz0.400m

5.3difference

path

m

4.1m0.12m13.4isdifferencepath

f

T T

f f T f

16.38: Solving Eq (16.17) for v, with v = 0, givesL

 25.0m s 775m s,Hz

1240Hz

1200

Hz1240

S L S

f v

or 780 m to two figures (the difference in frequency is known to only two figures) sNote that vS0, since the source is moving toward the listener

16.39: Redoing the calculation with +20.0 m for s vSand20.0m/sfor vL gives 267 Hz

16.40: a) From Eq 16.17, with vS0,vL 15.0m s, f A 375Hz

b) With vS 35.0m s, vL 15.0m s, f B 371Hz

c) f A  f B 4Hz(keepingan extrafigurein f A) The difference between the frequencies is known to only one figure

16.41: In terms of wavelength, Eq (16.29) is

v v

v v

a) 0, 25.0mand  344319 344m s 400Hz 0.798m

L S

these wavelengths are c) 431 Hz and d) 373 Hz

16.42: a) In terms of the period of the source, Eq (16.27) becomes

.sm25.0s1.6

m12.0sm32.0

S

T v v

b) Using the result of part (a) in Eq (16.18), or solving Eq (16.27) for v and S

substituting into Eq (16.28) (making sure to distinguish the symbols for the different wavelengths) gives 0.91m

v v

v f

b) vS0, vL v 2

kHz50.1

2

S 2

3 S

f

This is less than the answer in part (a)

The waves travel in air and what matters is the velocity of the listener or source relative to the air, not relative to each other

16.44: For a stationary source, S 0,so L S 1 L  S,

This is negative because the listener is moving away from the source

16.45: a) vL 18.0m/s,vS 30.0m s,and Eq.16.29gives  314362 262Hz

(sm344(70.1(

16.47: a) Mathematically, the waves given by Eq (16.1) and Eq (16.4) are out of

phase Physically, at a displacement node, the air is most compressed or rarefied on either side of the node, and the pressure gradient is zero Thus, displacement nodes are pressure antinodes b) (This is the same as Fig (16.3).) The solid curve is the pressure and the dashed curve is the displacement

c)

The pressure amplitude is not the same The pressure gradient is either zero or undefined At the places where the pressure gradient is undefined mathematically (the

“cusps” of the y - xplot), the particles go from moving at uniform speed in one direction

to moving at the same speed in the other direction In the limit that Fig (16.43) is an accurate depiction, this would happen in a vanishing small time, hence requiring a very large force, which would result from a very large pressure gradient d) The statement is true, but incomplete The pressure is indeed greatest where the displacement is zero, but the pressure is equal to its largest value at points other than those where the displacement

is zero

16.48: The altitude of the plane when it passes over the end of the runway is

m14515

m)tan 1200

m

(1740   , and so the sound intensity is 1 (1.45)2 of what the intensity would be at 100 m The intensity level is then

(1.45)  96.8dB,log

10dB0

so the airliner is not in violation of the ordinance

16.49: a) Combining Eq (16.14) and Eq (16.15),

0 max  2ρvI 10β  2(1.20kg m )(344m s)(10 W m )10

s)m344(Pa) 10144.1(2

9 5

2 max

v p Bk

p A

c) The distance is proportional to the reciprocal of the square root of the intensity, and hence to 10 raised to half of the sound intensity levels divided by 10 Specifically,

m

9.62m)10

00.5

16.50: a) 10( 10 dB)

I IA

sm0.3444

s 1 s

4 s 1 f 1 f

nf 3 gives s 4

Flute harmonic 3Nresonates with string harmonic 4N,N 1,3,5,

16.52: (a) The length of the string is d L10, so its third harmonic has frequency

s

μv F

L

(b) If the tension is doubled, all the frequencies of the string will increase by a factor of

2 In particular, the third harmonic of the string will no longer be in resonance with the first harmonic of the pipe because the frequencies will no longer match, so the sound produced by the instrument will be diminished

(c) The string will be in resonance with a standing wave in the pipe when their

frequencies are equal Using string

n the harmonics of the string are n3n3,9,15,

16.53: a) For an open pipe, the difference between successive frequencies is the

fundamental, in this case 392 Hz, and all frequencies would be integer multiples of this frequency This is not the case, so the pipe cannot be an open pipe For a stopped pipe, the difference between successive frequencies is twice the fundamental, and each

frequency is an odd integer multiple of the fundamental In this case,

.9Hz1764 ,7Hz1372 and

,

4

16.54: The steel rod has standing waves much like a pipe open at both ends, as shown

in Figure (16.18) An integral number of half wavelengths must fit on the rod, that is,

2L

nv

f n 

a) The ends of the rod are antinodes because the ends of the rod are free to ocsillate

b) The fundamental can be produced when the rod is held at the middle because a node is located there

1.50m 1980Hz.

2

sm59411

f

d) The next harmonic is n2,or f2 3961Hz.We would need to hold the rod at an

n = 2 node, which is located at L from either end, or at 0.375 m from either end.4

16.55: The shower stall can be modeled as a pipe closed at both ends, and hence there

are nodes at the two end walls Figure (15.23) shows standing waves on a string fixed at

both ends but the sequence of harmonics is the same, namely that an integral number of half wavelengths must fit in the stall

a) The condition for standing waves is nv L

232

116 1

Hz f( ) n

Note that the fundamental and second harmonic, which would have the greatest amplitude, are frequencies typically in the normal range of male singers Hence, men do sing better in the shower! (For a further discussion of resonance and the human voice, see

Thomas D Rossing , The Science of Sound, Second Edition, Addison-Wesley, 1990,

especially Chapters 4 and 17.)

16.56: a) The cross-section area of the string would be

,m101.29Pa)10(7.0

mkg10(7.8

kg)1000.4(

2 6 3

V L

b) Using the above result in Eq (16.35) gives f1 377Hz,or 380Hzto two figures

16.57: a) The second distance is midway between the first and third, and if there are no

other distances for which resonance occurs, the difference between the first and third positions is the wavelength 0.750m (This would give the first distance as

.39.1K)K)(350.15mol

J3145.8(

s)mmol)(375kg

108.28(γ

2 3

c) Since the first resonance should occur at τ 40.875mbut actually occurs at m,

18

0 the difference is 0.0075m

16.58: a) Considering the ear as a stopped pipe with the given length, the frequency of

the fundamental is f1 v 4L(344m s) (0.10m)3440Hz;3500Hzis near the

resonant frequency, and the ear will be sensitive to this frequency b) The next resonant frequency would be 10,500Hzand the ear would be sensitive to sounds with frequencies close to this value But 7000Hzis not a resonant frequency for an open pipe and the ear

is not sensitive at this frequency

16.59: a) From Eq (15.35), with m the mass of the string and M the suspended mass,

  

ρ L πd

Mg mL

)smkg)(9.8010

0.420(

3 3

2 2

6

2 3