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Unformatted text preview: or 263 Hz. Thus, the tuningfork
frequency must be 263 Hz . Chapter 17 Problems 917 22. SSM REASONING AND SOLUTION The first case requires that the frequency be either
440 Hz − 5 Hz = 435 Hz or 440 Hz + 5 Hz = 445 Hz The second case requires that the frequency be either
436 Hz − 9 Hz = 427 Hz or 436 Hz + 9 Hz = 445 Hz The frequency of the tuning fork is 445 Hz . 23. REASONING The beat frequency is the difference between two sound frequencies.
Therefore, the original frequency of the guitar string (before it was tightened) was either
3 Hz lower than that of the tuning fork (440.0 Hz − 3 Hz = 337 Hz) or 3 Hz higher (440.0
Hz + 3 Hz = 443 Hz):
443 Hz
440.0 Hz
437 Hz } 3Hz beat frequency
} 3Hz beat frequency To determine which of these frequencies is the correct one (437 or 443 Hz), we will use the
information that the beat frequency decreases when the guitar string is tightened
SOLUTION When the guitar string is tightened, its frequency of vibration (either 437 or
443 Hz) increases. As the drawing below shows, when the 437Hz frequency increases, it
becomes closer to 440.0 Hz, so the beat frequency decreases. When the 443Hz frequency
increases, it becomes farther from 440.0 Hz, so the beat frequency increases. Since the
problem states that the beat frequency decreases, the original frequency of the guitar string
was 437 Hz . } 443 Hz
440.0 Hz Beat frequency increases } Beat frequency decreases 437 Hz
Tuning
fork Original
string Tightened
string 24. REASONING The beat frequency heard by the bystander is the difference between the
frequency of the sound that the bystander hears from the moving car and that from the
(stationary) parked car. The bystander hears a frequency fo from the moving horn that is
greater than the emitted frequency fs. This is because of the Doppler effect (Section 16.9). 918 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA The bystander is the observer of the sound wave emitted by the horn. Since the horn is
moving toward the observer, more condensations and rarefactions of the wave arrive at the
observer’s ear per second than would otherwise be the case. More cycles per second means
that the observed frequency is greater than the emitted frequency.
The bystander also hears a frequency from the horn of the stationary car that is equal to the
frequency fs produced by the horn. Since this horn is stationary, there is no Doppler effect.
SOLUTION According to Equation 16.11, the frequency fo that the bystander hears from
the moving horn is
1
f o = fs v
1− s v where fs is the frequency of the sound emitted by the horn, vs is the speed of the moving
horn, and v is the speed of sound. The beat frequency heard by the bystander is fo – fs, so we
find that
1
1 f o − fs = fs − 1 − fs = fs v
v
1− s 1− s v
v 1 = ( 395 Hz ) − 1 = 14 Hz
12.0 m/s
1− 343 m/s 25. REASONING When the wave created by the tuning fork is superposed on the sound wave
traveling through the seawater, the beat frequency fbeat heard by the underwater swimmer is
equal to the difference between the two frequencies: fbeat = f wave − ffork (1) We know the frequency ffork of the tuning fork, and we will determine the frequency fwave of
the sound wave from its wavelength λ and speed v via v = f w ave λ (Equation 16.1). The
speed of a sound wave in a liquid is given by v = Bad ρ
adiabatic bulk modulus of the liquid, and ρ is its density. (Equation 16.6), where Bad is the SOLUTION Solving v = f w ave λ (Equation 16.1) for fwave yields f wave =
this result into Equation (1), we obtain v λ . Substituting Chapter 17 Problems f beat = f wave − f fork = Substituting v = Bad ρ v λ λ − f fork (2) (Equation 16.6) into Equation (2), we find that Bad
f beat = v 919 − f fork = ρ
− f fork =
λ 2.31×109 Pa
1025 kg/m3
− 440.0 Hz = 8 Hz
3.35 m 26. REASONING The blows of the carpenters’ hammers fall at frequencies f1 and f2, which are
related to the period T between blows by T = 1 f (Equation 10.5). The beat frequency
f beat = f1 − f 2 of their hammer blows is the frequency at which the blows fall at the same
instant. The time between these simultaneous blows is the period Tbeat of the beats.
SOLUTION
a. When the second carpenter hammers more rapidly than the first, the frequency of the
second carpenter’s hammer blows is larger than that of the first carpenter’s blows: f2 > f1.
Therefore, the beat frequency is given by
f beat = f1 − f 2 = f 2 − f1 (1) Solving Equation (1) for f2, we obtain
f 2 = f1 + f beat (2) The relation T = 1 f (Equation 10.5) tells us that the period T2 of the second carpenter’s
hammer blows is the reciprocal of the frequency f2. Therefore,
T2 = 1
1
=
f 2 f1 + f beat (3) Again invoking Equation 10.5, we can rewrite Equation (3) in terms of periods, rather than
frequencies:
T2 = 1
1
Tbeat + 1
T1 = 1
1
1
+
0.75 s 4.6 s = 0.64 s 920 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA b. If the second carpenter hammers less rapidly than the first carpenter,...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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