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

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

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,

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



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
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:




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








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(

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

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)

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

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

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
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


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





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



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

 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

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 


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