CHAPTER
16
WAVES AND SOUND
CONCEPTUAL QUESTIONS
____________________________________________________________________________________________
6
.
REASONING AND SOLUTION
A rope of mass
m
is hanging down from the ceiling.
Nothing is attached to the loose end of the rope.
A transverse wave is traveling up the rope.
The tension in the rope is not constant. The lower portion of the rope pulls down on the
higher portions of the rope. If we imagine that the rope is divided into small segments, we
see that the segments near the top of the rope are being pulled down by more weight than the
segments near the bottom. Therefore, the tension in the rope increases as we move up the
rope. The speed of a transverse wave on the rope is given by Equation 16.2:
wave
/(
/
)
v
F
m L
=
. From Equation 16.2 we see that, as the tension
F
in the rope increases,
the speed of the wave increases. Therefore, as the transverse wave travels up the rope, the
speed of the wave increases.
____________________________________________________________________________________________
9
.
REASONING AND SOLUTION
As the disturbance moves outward when a sound wave is
produced, it compresses the air directly in front of it. This compression causes the air
pressure to rise slightly, resulting in a condensation that travels outward. The condensation
is followed by a region of decreased pressure, called a rarefaction. Both the condensation
and rarefaction travel away from the speaker at the speed of sound. As the condensations
and rarefactions of the sound wave move away from the disturbance, the individual air
molecules are not carried with the wave. Each molecule executes simple harmonic motion
about a fixed equilibrium position. As the wave passes by, all the particles in the region of
the disturbance participate in this motion. There are no particles that are always at rest.
____________________________________________________________________________________________
10
.
REASONING AND SOLUTION
Assuming that we can treat air as an ideal gas, then the
speed of sound in air is given by Equation 16.5,
/
v
kT
m
γ
=
, where
γ
is the ratio of the
specific heats
c
P
/
c
V
,
k
is Boltzmann's constant,
T
is the Kelvin temperature, and
m
is the
mass of a molecule of the gas.
We can see from Equation 16.5 that the speed of sound in air is proportional to the square
root of the Kelvin temperature of the gas. Therefore, on a hot day, the speed of sound in air
is greater than it is on a cold day. Hence, we would expect an echo to return to us more
quickly on a hot day as compared to a cold day, other things being equal.
____________________________________________________________________________________________
14
.
REASONING AND SOLUTION
A source is emitting sound uniformly in all directions.
According to Equation 16.9,
I
=
P
/(4
π
r
2
)
, the intensity of such a source varies as 1/
r
2
.
Thus, the intensity
I
at a point in space depends on the distance of that point from the source.
A flat surface faces the source. As suggested in the following figure, the distance between
the source and the flat sheet varies, in general, from point to point on the sheet. The figure
indicates that, as we move up the screen, the distance between the source and the screen
increases. Therefore, the sound intensity is not the same at all points on the screen.
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 Spring '09
 DARICI
 Mass, Frequency, intensity level, sound intensity level

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