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5.10. According to Table 5.1, sound moves through air at 0°C with 5.16. The speed of sound at 0.0°C is 1,087 ft /s, and

a velocity of 331 m/s and through steel with a velocity of 2.00 ft /s
_

[

5,940 m/s. Th erefore, (a) v T F = v 0 + | °C | T P ]

d _

(a) v = 331 m/s v = ∴ d = vt 2.00 ft /s
t _

[0.0°C]

t = 8.00 s = 1,087 ft/s + | °C |

m _

d = ? = ( 331 ) (8.00 s) _
s
ft /s
m _ = 1,087 + (2.00)(0.0) ft /s + °C / × °C /

= 331 × 8.00 × s

s
= 1,087 ft/s + 0.0 ft/s
= 2,648 m
= 1,087 ft /s
= 2.65 km
2.00 ft /s
_
d _ (b) v 20° = 1,087 ft/s + | |
[20.0°C]
(b) v = 5,940 m/s v = ∴ d = vt °C

t
t = 8.00 s
m _

d = ? = ( 5,940 ) (8.00 s) = 1,087 ft /s + 40.0 ft /s
s
m _ = 1,127 ft /s

= 5,940 × 8.00 × s
s 2.00 ft /s
_

(c) v 40° = 1,087 ft/s + | | [40.0°C]

= 47,520 m °C
= 47.5 km = 1,087 ft /s + 80.0 ft /s
5.11. v = f λ = 1,167 ft /s
_
1 _ 2.00 ft /s

= (10 ) (0.50 m) (d) v 80° = 1,087 ft/s + | | [80.0°C]

s °C
m _ = 1,087 ft /s + 160 ft /s

= 5
s
= 1,247 ft /s
5.12. The distance between two consecutive condensations (or

rarefactions) is one wavelength, so λ = 3.00 m and 5.17. For consistency with the units of the equation given, 43.7°F is
first converted to 6.50°C. The velocity of sound in this air is

v = f λ
2.00 ft /s
_
1 _ = v 0 + | |

[T P ]

= (112.0 ) (3.00 m) v T F °C
s
m _ 2.00 ft /s
_
= 336 = 1,087 ft /s + | |

[6.50°C]
s °C
5.13. (a) One complete wave every 4.0 s means that T = 4.0 s. = 1,087 ft/s + 13.0 ft/s
1 _
(b) f = = 1,100 ft/s

T

1 _ The distance that a sound with this velocity travels in the given
=
4.0 s time is
d _

1 _ 1 _ v = ∴ d = vt
t
=

4.0 s = (1,100 ft/s)(4.80 s)
1 _ _
ft·s
= 0.25 = (1,100)(4.80)
s s
= 0.25 Hz = 5,280 ft
5,280 ft
_

5.14. The distance from one condensation to the next is one = 2
wavelength, so
= 2,640 ft
v _

v = f λ ∴ λ = Since the sound traveled from the rifle to the cliff and then
f
back, the cliff must be about one-half mile away.

m _

330 5.18. This problem requires three steps: (1) conversion of the °F

s
= _ temperature value to °C, (2) calculation of the velocity of sound

1 _
260
s in air at this temperature, and (3) calculation of the distance
_ s from the calculated velocity and the given time:
330 m _ _
= ×

260 s 1
2.00 ft /s
_

[

= 1.3 m v T F = v 0 + | °C | T P ]

1 _
_

5.15. (a) v = f λ = (256 ) (1.34 m) = 343 m/s 2.00 ft /s

s = 1,087 ft/s + | | [26.67°C]

1 _ °C
(b) = (440.0 ) (0.780 m) = 343 m/s

s = 1,087 ft /s + 53.0 ft/s = 1,140 ft /s
1 _ d _
(c) = (750.0 ) (0.457 m) = 343 m/s v = ∴ d = vt

s t
1 _ = (1,140 ft /s)(4.63 s)
(d) = (2,500.0 ) (0.137 m) = 343 m/s
s
= 5,280 ft (1 mile)

658 APPENDIX E Solutions for Group A Parallel Exercises E-16