 
Problem
1. Sllow that the quantity
has the units of
speed.
Solution
The units of pressure (force per unit area) divided by
3
density (mass per unit volome) are (~/m~)/(k~/m
)
=
(~/k~)(m~/m~)
=
(m/s2)m
=
(m/~)~,
or those of
speed squared.


Problem
9.
A
gas with density 1.0 kg/m3 and pressure
8.0x104 iV/m2 has sound speed 365 m/s. Are the
gas molecules monatomic or diatomic?
Solution
Solving for
7
in Equation 171, we find
7
=
pv2/P
=
(1.0 kg/m3)(365 m/s)*/(8.0~10'
~/m~)
=
1.67, very
close to the value for an ideal monatomic gas.
(Actually,
7

513
=
1.35~10~
for this gas.)
.
Problem
15. Sound intensity in normal conversation is about
1 pw/m2. What is the displacement amplitude of
air in a 2.5kHz sound wave with this intensity?
Solution
As in Example 172, Equation 173c, combined with
the atmospheric data in Example 171, can be used to
calculate the displacement amplitude for sound waves
of the specified frequency and intensity:
2(106 w/m2)
=
4.44 nm.
(1.20 kg/m3) (343 m/s)

Problem
18.
A
"tweeter" loudspeaker emitting 5.0 kHz sound
has an oscillation amplitude of 1
pm.
What must
be the oscillation amplitude of a "woofer" speaker
producing the same sound intensity at 30 Hz?
Solution
For the same
same air, Equation 173c
constant. Therefore, a
comparison of the woofer and tweeter yields (so),
=
'
(SO)~W*/W~
=
(1 pm)(5 kHz/30 Hz)
=
167 pm.
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 Spring '05
 wurthmeir
 Force, Wavelength, Fundamental frequency

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