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Unformatted text preview: initial (4) Substituting Equations (3) and (4) into Equation (1) gives as = vs, final − vs, initial
t fs fs v
− 1 − v −1 f o, final f o, initial v f =
s
= t
t 1
1
− f o, final f o, initial SOLUTION The acceleration of the car is v f
as = s
t 1
1 ( 343 m/s ) (1.00 ×103 Hz ) 1
1 2
−
− = 1.5 m/s
=
14.0 s f o, final fo, initial 912.0 Hz 966.0 Hz ______________________________________________________________________________ Chapter 16 Problems 887 86. REASONING Following its release from rest, the platform begins to move down the
incline, picking up speed as it goes. Thus, the platform’s velocity points down the incline,
and its magnitude increases with time. The reason for the increasing velocity is gravity. The
acceleration due to gravity points vertically downward and has a component along the
length of the incline. The changing velocity is related to the acceleration of the platform
according to Equation 2.4, which gives the acceleration as the change in the velocity divided
by the time interval during which the change occurs.
The frequency detected by the microphone at the instant the platform is released from rest is
the same as the frequency broadcast by the speaker. However, as the platform begins to
move away from the speaker, the microphone detects fewer wave cycles per second than the
speaker broadcasts. In other words, the microphone detects a frequency that is smaller than
that broadcast by the speaker. This is an example of the Doppler effect and occurs because
the platform is moving in the same direction as the sound is traveling. As the platform picks
up speed, the microphone detects an ever decreasing frequency. The speaker is the source of
the sound, and the microphone is the “observer.” Since the source is stationary and the
observer is moving away from the source, the Dopplershifted observed frequency is given
by Equation 16.14.
SOLUTION The acceleration a is directed down the incline and is the change in the
velocity divided by the time interval during which the change occurs. The change in the
velocity is the velocity vo,t at a later time t minus the velocity vo,t at an earlier time t0. Thus,
0 according to Equation 2.4, the acceleration is a= vo, t − vo, t 0 t − t0 Equation 16.14 gives the frequency fo detected by the microphone in terms of the frequency
fs emitted by the speaker, the speed vo, and the speed v of sound: v
f o = fs 1 − o v Solving for the speed of the mike gives f
vo = v 1 − o fs Using this result, we can determine the speed of the microphoneplatform at the two given
times:
9939 Hz [t = 1.5 s]
vo,1.5 s = ( 343 m/s ) 1 − = 2.1 m/s 1.000 ×104 Hz 888 WAVES AND SOUND [t = 3.5 s] 9857 Hz vo, 3.5 s = ( 343 m/s ) 1 − = 4.9 m/s 1.000 ×104 Hz Using these two values for the velocity, we can now obtain the acceleration using
Equation 2.4
vo, 3.5 s − vo,1.5 s 4.9 m/s − 2.1 m/s
a=
=
= 1.4 m/s 2
t − t0
3.5 s − 1.5 s 87. SSM REASONING
a. Since the two submarines are approaching each other head on, the frequency fo detected
by the observer (sub B) is related to the frequency fs emitted by the source (sub A) by
vo 1+ v
f o = fs 1 − vs v (16.15) where vo and vs are the speed of the observer and source, respectively, and v is the speed of
the underwater sound
b. The sound reflected from submarine B has the same frequency that it detects, namely, fo.
Now sub B becomes the source of sound and sub A is the observer. We can still use
Equation 16.15 to find the frequency detected by sub A. SOLUTION
a. The frequency fo detected by sub B is vo 1 + v
f o = fs vs 1−
v 8 m/s 1 + 1522 m/s = 1570 Hz = (1550 Hz ) 1 − 12 m/s 1522 m/s b. The sound reflected from submarine B has the same frequency that it detects, namely,
1570 Hz. Now sub B is the source of sound whose frequency is fs = 1570 Hz. The speed of
sub B is vs = 8 m/s. The frequency detected by sub A (whose speed is vo = 12 m/s) is
12 m/s vo 1 + 1522 m/s 1 + v = 1590 Hz
f o = fs = (1570 Hz ) 1 − 8 m/s 1 − vs 1522 m/s v ______________________________________________________________________________ Chapter 16 Problems 889 88. REASONING AND SOLUTION The maximum observed frequency is fomax, and the
minimum observed frequency is fomin. We are given that fomax – fomin = 2.1 Hz, where v
f omax = f s 1 + o v (16.13) v
f omin = fs 1 − o v (16.14) and We have v v
v f omax − f omin = f s 1 + o − fs 1 − o = 2 fs o v
v v
We can now solve for the maximum speed vo of the microphone: f omax − f omin
vo = v 2 fs 2.1 Hz = (343 m/s) = 0.82 m/s 2(440 Hz) Using vmax = vo = Aω, Equation 10.6, where A is the amplitude of the simple harmonic
2π
2π
motion and ω is the angular frequency, ω =
=
= 3.1 rad/s , we have
T
2.0 s
vo 0.82 m/s
= 0.26 m
ω 3.1 rad/s
______________________________________________________________________________
A= = 89. REASONING Knowing that the threshold of hearing corresponds to an intensity of
I0 = 1.00 × 10–12 W/m2, we can solve Equation 16.10 directly for the desired intensity.
SOLUTION Using Equation 16.10, we find I I0 β = 115 dB = (10 d...
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 Spring '13
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 Physics, The Lottery

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