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Chapter 17: Problem 80

A detector initially moves at constant velocity directly toward a stationary sound source and then (after passing it) directly from it. The emitted frequency is \(f .\) During the approach the detected frequency is \(f_{\text {app }}^{\prime}\) and during the recession it is \(f_{\text {rec }}^{\prime}\) If the frequencies are related by \(\left(f_{\text {app }}^{\prime}-f_{\text {rec }}^{\prime}\right) / f=0.500\), what is the ratio \(v_{D} / v\) of the speed of the detector to the speed of sound?

Short Answer

Expert verified
The ratio is \(\frac{v_D}{v} = 0.250\).

Step by step solution

01

Understand the Doppler Effect Formula for Approach

The Doppler effect describes how the observed frequency changes when there is relative motion between a sound source and a detector. For an approaching detector, the observed frequency is given by: \(f'_{\text{app}} = f \frac{v + v_D}{v}\) where \(v\) is the speed of sound, \(v_D\) is the speed of the detector, and \(f\) is the emitted frequency by the source.
02

Apply Doppler Effect Formula for Recession

Similarly, for a detector moving away from the sound source, the observed frequency is given by: \(f'_{\text{rec}} = f \frac{v - v_D}{v}\). Here, \(f'_{\text{rec}}\) is the detected frequency when the detector is moving away.
03

Calculate the Difference in Frequencies

According to the problem, the difference between the frequencies during approach and recession is given by: \(\frac{f'_{\text{app}} - f'_{\text{rec}}}{f} = 0.500\). We need to use the expressions for \(f'_{\text{app}}\) and \(f'_{\text{rec}}\) from the previous steps to compute this.
04

Plug in the Expressions and Simplify

Plug in \(f'_{\text{app}} = f \frac{v + v_D}{v}\) and \(f'_{\text{rec}} = f \frac{v - v_D}{v}\) into the given relationship: \(\frac{f \frac{v + v_D}{v} - f \frac{v - v_D}{v}}{f} = 0.500\). This simplifies to: \(\frac{2v_D}{v} = 0.500\).
05

Solve for the Ratio \(\frac{v_D}{v}\)

Rearrange the equation from the previous step: \(\frac{2v_D}{v} = 0.500\). Solving for the ratio, we divide both sides by 2: \(\frac{v_D}{v} = 0.250\). Thus, the ratio of the speed of the detector to the speed of sound is 0.250.

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Key Concepts

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

Frequency Shift
The frequency shift is a fundamental aspect of the Doppler Effect. It describes how the frequency of a sound wave changes due to the movement between the source and the observer.
When a sound source moves toward a detector, the waves are compressed, resulting in a higher observed frequency. This is called a positive frequency shift. Conversely, when the source moves away, the waves are stretched, leading to a lower frequency; this is a negative frequency shift. These changes are essential in numerous applications, such as radar and medical imaging.
Understanding the frequency shift is crucial for solving Doppler Effect problems. This is because it helps to quantify the difference in frequency perceived by the detector, both when it's moving toward and away from the source.
Sound Waves
Sound waves are longitudinal waves that travel through a medium鈥攍ike air, water, or solids. They compress and expand the particles in the medium, which we perceive as sound.
The speed of sound is affected by the medium through which it travels. For instance, sound moves faster in water than in air, and faster in solids than in liquids.
In the context of the Doppler Effect, understanding sound waves allows us to see how sound manifests itself as waves that can be compressed or stretched by motion. This change in the wave鈥檚 characteristics due to motion is what leads to frequency shifts, creating the Doppler Effect.
Relative Motion
Relative motion is a key concept in physics that refers to the movement of two objects with respect to each other. In the Doppler Effect, this relative motion is what causes the change in frequency that is observed.
Whether the detector is moving towards or away from the sound source affects the frequency detected. When the detector moves towards the source, the distance particles must travel is reduced, increasing frequency. Conversely, moving away increases the distance, thus lowering the frequency. This fundamental principle of relative motion is at the heart of the calculations for frequency changes in Doppler Effect scenarios.
Understanding relative motion helps explain why different frequencies are detected based on the direction and speed of the detector relative to the sound source. Keeping this principle in mind, we can calculate and predict the frequency changes that will be observed.

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Most popular questions from this chapter

Organ pipe \(A\), with both ends open, has a fundamental frequency of \(300 \mathrm{~Hz}\). The third harmonic of organ pipe \(B\), with one end open, has the same frequency as the second harmonic of pipe A. How long are (a) pipe \(A\) and (b) pipe \(B\) ?

An ambulance with a siren emitting a whine at \(1600 \mathrm{~Hz}\) overtakes and passes a cyclist pedaling a bike at \(2.44 \mathrm{~m} / \mathrm{s}\). After being passed, the cyclist hears a frequency of \(1590 \mathrm{~Hz} .\) How fast is the ambulance moving?

When you "crack" a knuckle, you suddenly widen the knuckle cavity, allowing more volume for the synovial fluid inside it and causing a gas bubble suddenly to appear in the fluid. The sudden production of the bubble, called "cavitation," produces a sound pulse-the cracking sound. Assume that the sound is transmitted uniformly in all directions and that it fully passes from the knuckle interior to the outside. If the pulse has a sound level of 62 \(\mathrm{dB}\) at your ear, estimate the rate at which energy is produced by the cavitation.

Approximately a third of people with normal hearing have ears that continuously emit a low-intensity sound outward through the ear canal. A person with such spontaneous otoacoustic emission is rarely aware of the sound, except perhaps in a noisefree environment, but occasionally the emission is loud enough to be heard by someone else nearby. In one observation, the sound wave had a frequency of \(1665 \mathrm{~Hz}\) and a pressure amplitude of \(1.13 \times 10^{-3} \mathrm{~Pa}\). What were (a) the displacement amplitude and (b) the intensity of the wave emitted by the ear?

\({ }^{\circ} 26\) A \(1.0 \mathrm{~W}\) point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a) \(1.0 \mathrm{~m}\) from the source and (b) \(2.5 \mathrm{~m}\) from the source.
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