/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 A siren emitting a sound of freq... [FREE SOLUTION] | 91Ó°ÊÓ

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A siren emitting a sound of frequency \(1000 \mathrm{~Hz}\) moves away from you toward the face of a cliff at a speed of \(10 \mathrm{~m} / \mathrm{s}\). Take the speed of sound in air as \(330 \mathrm{~m} / \mathrm{s}\). (a) What is the frequency of the sound you hear coming directly from the siren? (b) What is the frequency of the sound you hear reflected off the cliff? (c) What is the beat frequency between the two sounds? Is it perceptible (less than \(20 \mathrm{~Hz}\) )?

Short Answer

Expert verified
(a) 971 Hz, (b) 1000 Hz, (c) 29.4 Hz; perceptible.

Step by step solution

01

Understand the Doppler Effect Formula

The Doppler Effect formula for sound frequency when the source is moving away from the observer is given by: \[ f' = f \left( \frac{v}{v + v_s} \right) \] where \( f' \) is the observed frequency, \( f \) is the source frequency, \( v \) is the speed of sound in air, and \( v_s \) is the speed of the source relative to the medium.
02

Calculate Frequency from Siren

Substitute the given values into the formula: \[ f' = 1000 \left( \frac{330}{330 + 10} \right) \] Evaluate the expression: \( f' = 1000 \left( \frac{330}{340} \right) = 1000 \times 0.9706 \approx 970.6 \) Hz. So, the frequency you hear directly from the siren is approximately 971 Hz.
03

Calculate Frequency of Sound Reflected from Cliff

Consider that the sound reflects from the cliff and reaches back to you. Use the Doppler Effect again with the siren as an observer moving away from the source (reflection): \[ f'' = f' \left( \frac{v + v_o}{v} \right) = 970.6 \left( \frac{330 + 10}{330} \right) \] Evaluate the expression: \( f'' = 970.6 \times 1.0303 \approx 1000 \) Hz (back to the original since movement is now reversed). Thus, the frequency of sound you hear after reflection is approximately 1000 Hz.
04

Calculate the Beat Frequency

The beat frequency is given by the absolute difference between the two frequencies: \[ f_{beat} = |f'' - f'| = |1000 - 970.6| = 29.4 \] Hz. Thus, the beat frequency is 29.4 Hz.
05

Determine if Beat Frequency is Perceptible

To determine if the beat frequency of 29.4 Hz is perceptible, check if it is less than 20 Hz. Since \( 29.4 > 20 \), the beat frequency is not less than 20 Hz, meaning it is perceptible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency Calculation
The calculation of frequency when dealing with the Doppler Effect requires an understanding of how motion affects wave frequencies. The Doppler Effect describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source. In this situation, a siren is moving at a speed of \(10 \text{ m/s}\) towards a cliff while emitting a sound of \(1000 \text{ Hz}\). By using the Doppler Effect formula, we can calculate the observed frequency:

  • At first, we use the formula: \( f' = f \left( \frac{v}{v + v_s} \right) \).
  • We plug in the values: \( f = 1000 \text{ Hz}\), \( v = 330 \text{ m/s}\), \( v_s = 10 \text{ m/s}\).
  • This results in: \( f' = 1000 \left( \frac{330}{340} \right) \approx 970.6 \text{ Hz}\).
This calculated frequency displays how a moving source affects the observed frequency due to the Doppler Effect. This fundamental understanding is crucial whenever calculating frequencies in relative motion scenarios.
Beat Frequency
Beat frequency is an interesting phenomenon created when two sound waves of slightly different frequencies interfere with each other. In the context of this problem, once the siren's sound is reflected off the cliff, it produces two frequencies that listeners experience. The frequency of the reflected sound after recalculation returns to the original frequency of the siren, which is \(1000 \text{ Hz}\). The directly observed frequency was calculated to approximately \(971 \text{ Hz}\).

Calculating the beat frequency involves taking the absolute difference between the two frequencies:
  • The formula used is \( f_{\text{beat}} = |1000 - 970.6| \).
  • The result is a beat frequency of \(29.4 \text{ Hz}\).
This beat frequency indicates the rapid variation in intensity that a listener would perceive due to the interference of the two closely related frequencies. A perceptible beat frequency is typically below \(20 \text{ Hz}\), so \(29.4 \text{ Hz}\) is indeed perceptible as it exceeds the threshold. Recognizing and calculating beat frequencies help in understanding wave interference and sound perception in acoustics.
Sound Reflection
The process of sound reflection entails the redirection of sound waves when they hit a surface and bounce back. In the case of the problem where the siren faces a cliff, sound waves emit from the siren, travel through the air, strike the cliff, and reflect back.

Reflection affects how we perceive sound frequencies, especially in scenarios involving motion:
  • When the sound waves reflect off the cliff, they travel back towards the source, behaving almost like a new wave source.
  • The previously calculated frequency of \(970.6 \text{ Hz}\), once again passes through the Doppler Effect as it reflects, now with the siren acting as if it's moving towards the sound source.
  • As a result, the frequency returns to the original \(1000 \text{ Hz}\).
Understanding sound reflection is crucial in acoustics and various practical applications like acoustic engineering, sonar systems, and in environments where surface interactions alter sound experiences.

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Most popular questions from this chapter

A sperm whale (Fig. \(17-43 a\) ) vocalizes by producing a series of clicks. Actually, the whale makes only a single sound near the front of its head to start the series. Part of that sound then emerges from the head into the water to become the first click of the series. The rest of the sound travels backward through the spermaceti sac (a body of fat), reflects from the frontal sac (an air layer), and then travels forward through the spermaceti sac. When it reaches the distal sac (another air layer) at the front of the head, some of the sound escapes into the water to form the second click, and the rest is sent back through the spermaceti sac (and ends up forming later clicks). Figure \(17-43 b\) shows a strip-chart recording of a series of clicks. A unit time interval of \(1.0 \mathrm{~ms}\) is indicated on the chart. Assuming that the speed of sound in the spermaceti sac is \(1372 \mathrm{~m} / \mathrm{s}\), find the length of the spermaceti sac. From such a calculation, marine scientists estimate the length of a whale from its click series.

A sinusoidal sound wave moves at \(343 \mathrm{~m} / \mathrm{s}\) through air in the positive direction of an \(x\) axis. At one instant, air molecule \(A\) is at its maximum displacement in the negative direction of the axis while air molecule \(B\) is at its equilibrium position. The separation between those molecules is \(15.0 \mathrm{~cm}\), and the molecules between \(A\) and \(B\) have intermediate displacements in the negative direction of the axis. (a) What is the frequency of the sound wave? In a similar arrangement, for a different sinusoidal sound wave, air molecule \(C\) is at its maximum displacement in the positive direction while molecule \(D\) is at its maximum displacement in the negative direction. The separation between the molecules is again \(15.0 \mathrm{~cm}\), and the molecules between \(C\) and \(D\) have intermediate displacements. (b) What is the frequency of the sound wave?

A police car is chasing a speeding Porsche \(911 .\) Assume that the Porsche's maximum speed is \(80.0 \mathrm{~m} / \mathrm{s}\) and the police car's is \(54.0 \mathrm{~m} / \mathrm{s}\). At the moment both cars reach their maximum speed, what frequency will the Porsche driver hear if the frequency of the police car's siren is \(440 \mathrm{~Hz}\) ? Take the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\)

Two sound waves, from two different sources with the same frequency, \(540 \mathrm{~Hz}\), travel in the same direction at \(330 \mathrm{~m} / \mathrm{s}\). The sources are in phase. What is the phase difference of the waves at a point that is \(4.40 \mathrm{~m}\) from one source and \(4.00 \mathrm{~m}\) from the other?

A point source that is stationary on an \(x\) axis emits a sinusoidal sound wave at a frequency of \(686 \mathrm{~Hz}\) and speed \(343 \mathrm{~m} / \mathrm{s}\). The wave travels radially outward from the source, causing air molecules to oscillate radially inward and outward. Let us define a wavefront as a line that connects points where the air molecules have the maximum, radially outward displacement. At any given instant, the wavefronts are concentric circles that are centered on the source. (a) Along \(x\), what is the adjacent wavefront separation? Next, the source moves along \(x\) at a speed of \(110 \mathrm{~m} / \mathrm{s}\). Along \(x\), what are the wavefront separations (b) in front of and (c) behind the source?

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