Week 12 homework IMPORTANT NOTE ABOUT WEBASSIGN:

Week 12 homework
IMPORTANT NOTE ABOUT WEBASSIGN:
In the WebAssign versions of these problems, various details have been changed, so that
the answers will come out differently. The method to find the solution is the same, but
you will need to repeat part of the calculation to find out what your answer should have
been.
WebAssign Problem 1: A longitudinal wave with a frequency of 3.0 Hz takes 1.7 s to
travel the length of a 2.5-m Slinky (see Figure 16.3). Determine the wavelength of
the wave.
REASONING AND SOLUTION From Equation 16.1, we have λ = v/f. But v = x/t, so
we find
v
x
2.5 m
λ = =
=
= 0.49 m
f t f ( 1.7 s ) ( 3.0 Hz )
WebAssign Problem 2: Tsunamis are fast-moving waves often generated by underwater
earthquakes. In the deep ocean their amplitude is barely noticeable, but upon
reaching shore, they can rise up to the astonishing height of a six-story building. One
tsunami, generated off the Aleutian islands in Alaska, had a wavelength of 750 km
and traveled a distance of 3700 km in 5.3 h. (a) What was the speed (in m/s) of the
wave? For reference, the speed of a 747 jetliner is about 250 m/s. Find the wave’s
(b) frequency and (c) period.
REASONING The speed of a Tsunamis is equal to the distance x it travels divided by
the time t it takes for the wave to travel that distance. The frequency f of the wave is
equal to its speed divided by the wavelength λ, f = v/λ (Equation 16.1). The period T of
the wave is related to its frequency by Equation 10.5, T = 1/f.
SOLUTION
a.
The speed of the wave is (in m/s)
v=
b.
x 3700 × 103 m  1 h 
=

 = 190 m/s
t
5.3 h
 3600 s 
The frequency of the wave is
f =
c.
v
190 m/s
=
= 2.5 × 10− 4 Hz
λ 750 × 103 m
The period of any wave is the reciprocal of its frequency:
(16.1)
T=
1
1
=
= 4.0 × 103 s
−
4
f 2.5 × 10 Hz
(10.5)
WebAssign Problem 3: A steel cable of cross-sectional area
is kept
under a tension of
. The density of steel is 7860 kg/m3 (this is not the
linear density). At what speed does a transverse wave move along the cable?
REASONING AND SOLUTION The speed of the wave on the cable is
v=
F
=
(m / L)
F
=
ρA
1.00 × 104 N
= 21.2 m/s
(7860 kg/m3 )(2.83 × 10− 3 m 2 )
WebAssign Problem 4: The drawing shows two graphs that represent a transverse wave
on a string. The wave is moving in the + x direction. Using the information contained in
these graphs, write the mathematical expression (similar to Equation 16.3 or 16.4) for the
wave.
REASONING AND SOLUTION We find from the graph on the left that λ = 0.060 m –
0.020 m = 0.040 m and Α = 0.010 m. From the graph on the right we find that Τ = 0.30 s
– 0.10 s = 0.20 s. Then, f = 1/(0.20 s) = 5.0 Hz. Substituting these into Equation 16.3 we
get
2π x 

y = A sin  2π f t –

λ 

and
y = ( 0.010 m ) sin ( 10π t – 50π x )
WebAssign Problem 5: A monatomic ideal gas (γ = 1.67) is contained within a box
whose volume is 2.5 m3. The pressure of the gas is
. The total mass of
the gas is 2.3 kg. Find the speed of sound in the gas.
REASONING AND SOLUTION The speed of sound in an ideal gas is given by text
Equation 16.5:
γ kT
v=
m
(16.2)
where γ = cp/cv, k is Boltzmann's constant, T is the Kelvin temperature of the gas,
and m is the mass of a single gas molecule. If mTOTAL is the mass of the gas sample and N
is the total number of molecules in the sample, then the above equation can be written as
v=
γ kT
(mTOTAL / N )
=
γ N kT
mTOTAL
For an ideal gas, PV = NkT, so that Equation (1) becomes,
v=
γ PV
(1.67)(3.5×105 Pa)(2.5 m3 )
=
= 8.0 ×102 m/s
mTOTAL
2.3 kg
WebAssign Problem 6: Suppose that sound is emitted uniformly in all directions by a
public address system. The intensity at a location 22 m away from the sound source
is
. What is the intensity at a spot that is 78 m away?
REASONING AND SOLUTION The intensity of the sound falls off according to the
square of the distance according to Equation 16.9. Therefore, the intensity at the new
location will be
I 78
 22 m 
= 

 78 m 
2
( 3.0 × 10− 4 W/m2 ) =
2.4 × 10 –5 W/m 2
WebAssign Problem 7: You are riding your bicycle directly away from a stationary
source of sound and hear a frequency that is 1.0% lower than the emitted frequency.
The speed of sound is 343 m/s. What is your speed?
REASONING You hear a frequency fo that is 1.0% lower than the frequency fs emitted
by the source. This means that the frequency you observe is 99.0% of the emitted
frequency, so that fo = 0.990 fs. You are an observer who is moving away from a
stationary source of sound. Therefore, the Doppler-shifted frequency that you observe
is specified by Equation 16.14, which can be solved for the bicycle speed vo.
SOLUTION Equation 16.14, in which v denotes the speed of sound, states that
v 

f o = fs  1 − o 
v 

Solving for vo and using the fact that fo = 0.990 fs reveal that
(1)


f 
0.990 fs 
vo = v  1 − o  = ( 343 m/s )  1 −
 = 3.4 m/s



f
f
s 
s



WebAssign Problem 8: Interactive LearningWare 16.2 at
www.wiley.com/college/cutnell provides some pertinent background for this
problem. A convertible moves toward you and then passes you; all the while, its
loudspeakers are producing a sound. The speed of the car is a constant 9.00 m/s, and
the speed of sound is 343 m/s. What is the ratio of the frequency you hear while the
car is approaching to the frequency you hear while the car is moving away?
REASONING This problem deals with the Doppler effect in a situation where the source
of the sound is moving and the observer is stationary. Thus, the observed frequency
is given by Equation 16.11 when the car is approaching the observer and
Equation 16.12 when the car is moving away from the observer. These equations
relate the frequency fo heard by the observer to the frequency fs emitted by the source,
the speed vs of the source, and the speed v of sound. They can be used directly to
calculate the desired ratio of the observed frequencies. We note that no information is
given about the frequency emitted by the source. We will see, however, that none is
needed, since fs will be eliminated algebraically from the solution.
SOLUTION Equations 16.11 and 16.12 are

1 
f oApproach = fs 
 1 − v / v 
s


(16.11)

1 
f oRecede = fs 
 1 + v / v 
s


(16.12)
The ratio is
f oApproach
f oRecede

fs 
 1−
=

fs 
 1+
1 

vs / v 
9.00 m/s
343 m/s = 1.054
=
=
9.00
m/s
1  1 − vs / v 1 −

343 m/s
vs / v 
1 + vs / v
1+
As mentioned in the REASONING, the unknown source frequency fs has been
eliminated algebraically from this calculation.
WebAssign Problem 9: A 3.49-rad/s (
rpm) record has a 5.00-kHz tone cut in the
groove. If the groove is located 0.100 m from the center of the record (see drawing),
what is the wavelength in the groove?
REASONING AND SOLUTION First find the speed of the record at a distance of 0.100
m from the center:
(8.9)
v = rω = (0.100 m)(3.49 rad/s) = 0.349 m/s
The wavelength is, then,
–5
λ = v/f = (0.349 m/s)/(5.00 × 103 Hz) = 6.98 × 10 m
(16.1)
Practice conceptual problems:
6. 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 on the rope. As the wave travels up the
rope, does the speed of the wave increase, decrease, or remain the same? Give a reason
for your choice.
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: vwave = 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.
13. Some animals rely on an acute sense of hearing for survival, and the visible part of
the ear on such animals is often relatively large. Explain how this anatomical feature
helps to increase the sensitivity of the animal’s hearing for low-intensity sounds.
REASONING AND SOLUTION Animals that rely on an acute sense of hearing for
survival often have relatively large external ear parts. Sound intensity is defined as
the sound power P that passes perpendicularly through a surface divided by the area
A of that surface: I = P / A . For low intensity sounds, the power per unit area is
small. Relatively large outer ears have a greater area than smaller outer ears. Hence,
large outer ears intercept and direct more sound power into the auditory system than
smaller outer ears do.
17. A source of sound produces the same frequency underwater as it does in air. This
source has the same velocity in air as it does underwater. The observer of the sound is
stationary, both in air and underwater. Is the Doppler effect greater in air or underwater
when the source (a) approaches and (b) moves away from the observer? Explain.
REASONING AND SOLUTION According to Table 16.1, the speed of sound in air at
20 °C is v = 343 m/s, while its value in water at the same temperature is v = 1482
m/s. These values influence the Doppler effect, because of which an observer hears
a frequency fo that is different from the frequency fs that is emitted by the source of
sound. For purposes of this question, we assume that fs = 1000 Hz and that the speed
at which the source moves is vs = 25 m/s. Our conclusions, however, will be valid
for any values of fs and vs .
a. The Doppler-shifted frequency when the source approaches the observer is given
by Equation 16.11 as f o = fs  1 − (vs / v)  . Applying this equation for air and
water, we find
Air
fo =
1000 Hz
= 1079 Hz
1 − (25 m s )/(343 m s )
Water
fo =
1000 Hz
= 1017 Hz
1 − (25 m s )/(1482 m s )
The change in frequency due to the Doppler effect in air is greater.
b. The Doppler-shifted frequency when the source moves away from the observer is
given by Equation 16.12 as f o = fs  1 + (vs / v)  . Applying this equation for air
and water, we find
1000 Hz
fo =
= 932 Hz
Air
1 + (25 m s )/(343 m s )
Water
fo =
1000 Hz
= 983 Hz
1 + (25 m s )/(1482 m s )
The change in frequency due to the Doppler effect in air is greater.
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