/*! 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) \(970.59 \text{ Hz}\); (b) \(1031.25 \text{ Hz}\); (c) \(60.66 \text{ Hz}\), not perceptible.

Step by step solution

01

Identify the Parameters

We need to identify the known parameters in the problem:- Frequency of the siren (source), \( f_s = 1000 \text{ Hz} \).- Speed of the source, \( v_s = 10 \text{ m/s} \).- Speed of sound in air, \( v = 330 \text{ m/s} \).
02

Apply the Doppler Effect for the Direct Sound

The formula for the observed frequency when the source is moving away from the observer is:\[ f' = \left(\frac{v}{v + v_s}\right) f_s\]Substitute the known values:\[ f' = \left(\frac{330}{330 + 10}\right) \times 1000 = \left(\frac{330}{340}\right) \times 1000 = 970.59 \text{ Hz}\]The frequency of the sound you hear directly from the siren is approximately \(970.59 \text{ Hz}\).
03

Calculate the Frequency of the Reflected Sound

When the sound waves reflect off the cliff, the cliff acts as a stationary source emitting the frequency it receives. First, calculate the frequency the cliff 'hears' from the siren:\[ f'' = \left(\frac{v}{v - v_s}\right) f_s\]Substitute the known values:\[ f'' = \left(\frac{330}{330 - 10}\right) \times 1000 = \left(\frac{330}{320}\right) \times 1000 = 1031.25 \text{ Hz}\]So, the frequency reflected back is also \(1031.25 \text{ Hz}\).
04

Calculate the Beat Frequency

The beat frequency \( f_{beat} \) is given by the absolute difference between the two frequencies:\[ f_{beat} = \left| f' - f'' \right| = \left| 970.59 - 1031.25 \right| = 60.66 \text{ Hz}\]Thus, the beat frequency is approximately \(60.66 \text{ Hz}\).
05

Determine Perceptibility of the Beat Frequency

Beat frequencies are perceptible if they are less than \(20 \text{ Hz}\). Since \(60.66 \text{ Hz}\) is greater than \(20 \text{ Hz}\), this beat frequency is not perceptible.

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Key Concepts

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

Frequency calculation
When it comes to understanding sound waves, frequency is a fundamental concept. Frequency refers to how many times the sound wave repeats itself in one second. It's measured in Hertz (Hz). The higher the frequency, the higher the pitch of the sound we perceive.
In the exercise, we calculate the frequency using the Doppler Effect, which describes how the frequency of a wave changes when the source or listener is moving. The formula used in the exercise is a specific application of this principle, where the source is moving away from the observer. The frequency you hear, also known as the observed frequency \( f' \), is reduced compared to the source frequency \( f_s \). This is calculated as \( f' = \left(\frac{v}{v + v_s}\right) f_s \), where \( v \) is the speed of sound, and \( v_s \) is the speed of the source.
This results in a frequency lower than the original, as observed in the exercise solution (\(970.59 \text{ Hz}\)). It’s important to remember that motion affects how we perceive the frequency, showcasing how dynamic the interactions of waves can be.
Sound waves
Sound waves are vibrations that travel through the air (or another medium) and are detected by the ears. They are longitudinal waves, which means the vibrations occur in the same direction as the wave travels.
In physics, sound waves are typically studied in terms of their speed, frequency, and wavelength. The speed of sound in air, as mentioned in the exercise, is approximately \(330 \text{ m/s}\) under typical conditions. This speed can vary based on the medium and temperature.
The properties of waves like frequency and speed are crucial in determining how sound is transmitted and received. When a wave source moves relative to an observer, the Doppler Effect comes into play. This effect explains the changes in wave frequency experienced. When the source moves towards an observer, the waves seem to "bunch up," leading to a higher frequency. Conversely, as in the given problem, moving away stretches the waves apart, reducing the frequency as presented.
Beat frequency
Beat frequency is an interesting phenomenon occurring when two sound waves of slightly different frequencies interfere with each other. You hear this as a pulsing effect or "beats," which is due to the periodic variation in sound intensity.
The beat frequency is calculated as the absolute difference between the two frequencies involved. In our exercise, two sound waves are involved: one directly from the siren and one reflected off the cliff. These two create a beat frequency computed as \( f_{beat} = |f' - f''| \).
It is significant because it tells us how noticeable the differences in frequencies are. If the beat frequency is less than \(20 \text{ Hz}\), it is usually perceptible to human ears. In this exercise, a beat frequency of approximately \(60.66 \text{ Hz}\) was computed, indicating the beats are not likely perceptible as individual sounds due to this frequency exceeding the perceptual threshold.

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