Tài liệu Physics exercises_solution: Chapter 16 ppt

16.1: a)
 Appmfv then,1000if)b.344.0)Hz100(s)m344(λ
0
0
1000A
Therefore, the amplitude is
m.102.1
5

maxmax
increasing,Sincec) pBkAp 
while keeping A constant requires decreasing k, and increasing
π
, by the same factor.
Therefore the new wavelength is
Hz.50m,9.6)20(m)688.0(
m9.6
sm344
new
 f
16.2:
m.1021.3or ,
12
Hz)1000(Pa)
9
102.2(2
)sm1480(Pa)
2
100.3(
2
max






 AA
π
πBf
vp
The much higher bulk modulus
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),
.2λ2
max
vfπBABAπBkAp 
a)
Pa.7.78s)m(344Hz)150(m)1000.2(Pa)1042.1(2
55


π
b)
Pa.778Pa7.78100c) Pa.77.8Pa78.710 
The amplitude at
Hz1500
exceeds the pain threshold, and at
Hz000,15
the sound

would be unbearable.
16.4: The values from Example 16.8 are
Hz,1000Pa,1016.3
4
 fB
m.102.1
8
A
Using Example 16.5,
,sm295sm344
293K
K216
v
so the pressure
amplitude of this wave is
Pa).1016.3(
2
4
max
 A
v
πf
BBkAp
Pa.108.1m)102.1(
sm295
Hz)(10002
38 

π
This is

0.27Pa)10(3.0Pa)101.8(
23


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
22
s)400m)(8(so,)λ( BfρvB
Pa.1033.1)mkg1300(
103

b) Using Equation (16.8),
 
2
422
s10(3.9m)5.1()(

 ρtLρvY
Pa.1047.9)mkg6400(
103

16.6: a) The time for the wave to travel to Caracas was
s 579s 39min 9 
and the
speed was
sm10085.1

4

(keeping an extra figure). Similarly, the time for the wave to
travel to Kevo was 680 s for a speed of
s,m10278.1
4

and the time to travel to Vienna
was 767 s for a speed of
s.m10258.1
4

The average speed for these three
measurements is
s.m1021.1
4

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),
,
2

vB 
and using the given value of
33
mkg103.3 

and the
speeds found in part (a), the values for the bulk modulus are, respectively,

Pa.102.5 and Pa105.4 Pa,109.3
111111

These are larger, by a factor of 2 or 3, than
the largest values in Table (11-1).
16.7: Use
sm1482
water
v
at
C,20
as given in Table
 
1.16
The sound wave travels
in water for the same time as the wave travels a distance
m8.20m20.1m0.22 
in air,
and so the depth of the diver is
   
m.6.89
sm344
sm1482
m8.20m8.20
air
water

v
v
This is the depth of the diver; the distance from the horn is

m.8.90
16.8: a), b), c) Using Eq.
 
,10.16
   
 
sm1032.1
molkg1002.2
K15.300KmolJ3145.841.1
3
3
2
H





v
   
 
sm1002.1
molkg1000.4
K15.300KmolJ3145.867.1
3
3
e
H






v

   
 
.sm323
molkg109.39
K15.300KmolJ3145.867.1
3
Ar





v
d) Repeating the calculation of Example 16.5 at
K15.300T
gives
,sm348
air
v
and
so
airHeairH
94.2,80.3
2
vvvv 

and
.928.0
airAr
vv 
16.9: Solving Eq. (16.10) for the temperature,
  
K,191
KmolJ3145.840.1
hrkm6.3
sm1
0.85
hkm850
mol)kg108.28(
2
3
2
























R
Mv
T

or
C.82
b) See the results of Problem 18.88, the variation of atmospheric pressure
with altitude, assuming a non-constant temperature. If we know the altitude we can use
the result of Problem 18.88,
.1
0
0


















R
Mg
T
y
pp
Since
,yTT
o

m,C106.K,191for
2


T
and
).ft.840,44(m667,13K,273
0
 yT
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:

,
K273
m)(13,667)mC106(.
1p
m)C10K)(.6molJ315.8(
)smmol)(9.8kg108.28(
2
o
2
23
























p
and
,p13.)70(.p
o
66.5
o
p
or about .13 atm. Using an altitude of 13,667 m in the
equation derived in Example 18.4 gives
,p18.
o
p
which overestimates the pressure due
to the assumption of an isothermal atmosphere.
16.10: As in Example
K,294.15C21 with 5,-16 T
s.m80.344
molkg108.28
K)K)(294.15molJ3145.8)(04.1(
3








M
RT
v
The same calculation with
s,m344.22gives283.15T
so the increase is
s.m58.0
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(
m0.80
sm344
m0.80
311
Brassair



v
L
v
L
t
16.12:
mol)kg108.28(
K)K)(260.15molJ3145.8)(40.1(
mol)kg108.28(
K)K)(300.15molJ3145.8)(40.1(

33 






s.m24
(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
,by AρμA 
so combining Eq. (15.13) and Eq. (16.8) with the given
relations between the speeds,

900
so 900
Υ
AF
Αρ
F
ρ
Υ
16.14: a)
m.0.16
Hz220
)mkg10(8.9Pa)100.11(

3310



f
ρΥ
f
v
b) Solving for the amplitude A (as opposed to the area
)
2
πra 
in terms of the average
power
,
av
ΙaP 

2
av
)2(
ρΥω
aP
A


m.1029.3
))Hz220(2(Pa)100.11)(mkg109.8(
)m)10(0.800(W))1050.6(2
8

210
3
3
2-26






π
π
c)
s.m104.55m)10289.3)(Hz220(22
58 
 πΑfπω
16.15: a) See Exercise 16.14. The amplitude is

2
2
ρΒω
Ι


m.1044.9
Hz))3400(2(Pa)1018.2)(mkg1000(
)mW1000.3(2
11
293
26







π
The wavelength is
m.434.0
Hz3400
)mkg(1000Pa)1018.2(
39



f
ρB
f
v
b) Repeating the above with
Pa1040.1γ
5
 pB
and the density of air gives
m.0.100andm1066.5
9


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.
16.16: From Eq. (16.13),
,2
2
max
BvpI 
and from Eq. (19.21),
.
2

Bv 
Using
Eq. (16.7) to eliminate
 
.22,
2
max
2
max
ρBpBpρBIv 
Using Eq. (16.7) to
eliminate B,
.2)(2
2
max
22
max
ρvpρvvpI 
16.17: a)

Pa.95.1
s)m(344
m)10(5.00Hz)(150Pa)1042.1(22
max
65


π
v
πBfA
BkAp
b) From Eq. (16.14),
.mW104.58s))m344)(mkg2.1(2(Pa)95.1(2
23322
max

 ρvpI
c)
 
dB.6.96log10
12
3
10
104.58




16.18: (a) The sound level is
dB.57or ,logdB)(10so,logdB)10(

212
2
0
mW10
mW0.500
I
I


βββ
μ
b) First find v, the speed of sound at
C,0.20 
from Table 16.1,
s.m344v
The density of air at that temperature is
.mkg20.1
3
Using Equation (16.14),
.mW1073.2or ,
s)m344)(mkg20.1(2
)mN150.0(
2
25
3
222
max

 I
ρv

p
I
Using this in Equation
(16.15),
dB.74.4or ,
mW10
mW102.73
logdB)10(
212
25





ββ
16.19: a) As in Example 16.6,
dB.40.6.mW104.4
212
s)m344)(mkg20.1(2
Pa)100.6(
3
25




βI
16.20: a)
 

dB.0.6log10
4

I
I
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
25
1
5


of the intensity that he hears, and the difference in sound intensity levels is
dB.14log(25)10 
16.22:
dB25dB90dB75level)(Sound 
i
f
I
I
I
I
I
I
f
0
i
0

log10log10log10level)(Sound 
Therefore
i
I
I
f
log10dB25 
35.2
102.310
i
f


I
I
16.23:
,0.20Thus,.dB)log(10dB13or ,dB)log10(
0
00
 IIβ
I
I
I
I
or the intensity has
increased a factor of 20.0.
16.24: Open Pipe:
Hz594
2
1

1
v
f
v
L

Closed at one end:
f
v
L
 4
1
Taking ratios:

Hz297
2
Hz594
Hz594
4
2


f
fv
v
L
L
16.25: a) Refer to Fig. (16.18). i) The fundamental has a displacement node at
m600.0
2


L
, the first overtone mode has displacement nodes at
m300.0
4

L
m900.0and
4
3

L
and the second overtone mode has displacement nodes at
m000.1andm600.0m,200.0
6
5
26

L
LL
. ii) Fundamental:
0,:First m.200.1,0 L
m.200.1m,600.0
2
 L
L
m.200.1m,800.0m,400.00,:Second
3
2
3

 L
LL
b) Refer to Fig. (16.19); distances are measured from the right end of the pipe in
the figure. Pressure nodes at: Fundamental:
m200.1L
. First overtone:
m.200.1m,400.03  LL
Second overtone:
m, 240.05 L
m.200.1,m720.053  LL
Displacement nodes at Fundamental:
.0
First
overtone:
m.800.032,0 L
Second overtone:
m, 480.052 ,0 L m960.054 L
16.26: a)
Hz,382
m)450.0(2
)344(
2
1

sm
L
v
f
Hz,11473Hz,7642
131

 fff
Hz.15294
14
 ff
b)
Hz.13387Hz,9565Hz,5733Hz,191
171513
4
1
 fffffff
L
v
Note that
the symbol “
1
f
” denotes different frequencies in the two parts. The frequencies are not
always exact multiples of the fundamental, due to rounding.
c) Open:
,3.52
1
000,20

f
so the 52
nd
harmonic is heard. Stopped;
,7.104
1
000,20


f
so 103 rd
highest harmonic heard.
16.27:
Hz.25295 Hz,15173Hz,506
1312
m)4(0.17
m/s)344(
1
 fffff
16.28: a) The fundamental frequency is proportional to the square root of the ratio
M

(see Eq. (16.10)), so
Hz,767
00.4
8.28
)57(
3)5(
Hz)262(
He
air
air
He
airHe





M
M
ff
b) No; for a fixed wavelength , the frequency is proportional to the speed of sound in
the gas.
16.29: a) For a stopped pipe, the wavelength of the fundamental standing wave is
,m56.04 L
and so the frequency is
 
kHz.0.614m)56.0(sm344
1
f
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
 
,,33.15
2L
nv
n
f 
is useful. It is
important to remember the second overtone is the third harmonic. Solving for v,
,
2
n
Lf
n
v 

and inserting the data,
   
3
/s588m635.2
v
, and
.sm249v
16.31: a) For constructive interference, the path difference
m00.2d
must be equal to
an integer multiple of the wavelength, so
,λ
n
nd
 
.Hz172
m2.00
sm344
λ
n
nn
d
v
n
d
vnv
f
n









Therefore, the lowest frequency is 172 Hz.
b) Repeating the above with the path difference an odd multiple of half a
wavelength,
 
 
.Hz172
2
1
 nf
n
Therefore, the lowest frequency is
 
.0nHz86 
16.32: The difference in path length is
   
.2or ,2 xLxxLxxLx 
For
destructive interference,
))λ21((  nx
,and for constructive interference,
λ.nx 
The
wavelength is
 

m670.1)Hz206(sm344λ  fv
(keeping an extra figure), and so
to have
34,0  nLx
for destructive interference and
44  n
for constructive
interference. Note that neither speaker is at a point of constructive or destructive
interference.
a) The points of destructive interference would be at
m.1.42m,58.0x
b) Constructive interference would be at the points
m.1.83 m,1.00 m,17.0x
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:
   
m500.0Hz688sm344λ  fv
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
x
and the difference to A gets longer by

x
so the path difference
changes by 2x.

2λ2 x
and
m0.1254λ x
16.34: We are to assume
   
.m00.2Hz172m/s344λso,sm344  fvv
If
mr
A
00.8
and
B
r
are the distances of the person from each speaker, the condition for
destructive interference is
 
λ,
2
1
 nrr
AB
where n is any integer. Requiring
 
0λ
2
1

 nrr
AB
gives
   
,4m00.2m00.8
2
1

A
rn
so the smallest
value of
B
r
occurs when
,4n
and the closest distance to B is
 
 
m.00.1m00.24-m00.8
2
1

B
r
16.35:
   
m400.0Hz860sm344  fv

5.3

differencepath
m.4.1m0.12m13.4isdifferencepath The



The path difference is a half-integer number of wavelengths, so the interference is
destructive.
16.36:
,Sincea)
beat ba
fff 
the possible frequencies are 440.0
 Hz5.1Hz

Hz441.5or Hz5.438
b) The tension is proportional to the square of the frequency.
Therefore
 
Hz440
Hz5.12
2
2
i).So .2and 



T
T
f
f

T
T
ffTfT
.1082.6
3

 
.1082.6ii)
3
Hz440
Hz5.12




T
T
16.37: a) A frequency of
 
Hz110Hz112Hz108
2
1

will be heard, with a beat
frequency of 112 Hz–108 Hz = 4 beats per second. b) The maximum amplitude is the
sum of the amplitudes of the individual waves,
 
.m100.3m105.12
88 


The
minimum amplitude is the difference, zero.
16.38: Solving Eq. (16.17) for v, with
L
v
= 0, gives
 
,sm775sm0.25
Hz1240Hz1200
Hz1240
S
LS
L











v
ff
f
v
or 780
sm

to two figures (the difference in frequency is known to only two figures).
Note that
,0
S
v
since the source is moving toward the listener.
16.39: Redoing the calculation with +20.0
sm
for
LS
for m/s0.20and vv 
gives 267
Hz.
16.40: a) From Eq.
 
Hz.375,sm0.15,0 with ,17.16
LS



A
fvv
b)
Hz.371,sm0.15 ,sm35.0 With
LS



B
fvv

c)
)in figureextraan (keepingHz4
ABA
fff





. The difference between the
frequencies is known to only one figure.