/*! 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 80 A detector initially moves at co... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \( \frac{v_D}{v} \) is 0.250.

Step by step solution

01

Understand the Doppler Effect Equations

The Doppler effect describes how the frequency of waves changes relative to an observer when there is motion between the source and the observer. The frequency detected by the observer when moving towards the source is given by \( f_{\text{app}}' = f \left( \frac{v + v_D}{v} \right) \), and the frequency when moving away is \( f_{\text{rec}}' = f \left( \frac{v - v_D}{v} \right) \), where \( v \) is the speed of sound, \( v_D \) is the speed of the detector, and \( f \) is the emitted frequency.
02

Write the Given Condition

According to the problem, \( \frac{f_{\text{app}}' - f_{\text{rec}}'}{f} = 0.500 \). This condition describes the relationship between the changes in the detected frequencies during approach and recession.
03

Express the Detected Frequencies in Terms of Given Quantities

Substitute the expressions for \( f_{\text{app}}' \) and \( f_{\text{rec}}' \) into the condition: \[ \frac{f \left( \frac{v + v_D}{v} \right) - f \left( \frac{v - v_D}{v} \right)}{f} = 0.500 \]. Simplifying this gives \[ \frac{(v + v_D) - (v - v_D)}{v} = 0.500 \].
04

Simplify the Equation

The equation becomes \[ \frac{2v_D}{v} = 0.500 \]. Simplifying further gives \( 2 \frac{v_D}{v} = 0.500 \).
05

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

Divide both sides by 2 to isolate \( \frac{v_D}{v} \): \[ \frac{v_D}{v} = \frac{0.500}{2} = 0.250 \].
06

Conclusion

The ratio \( \frac{v_D}{v} \) of the speed of the detector to the speed of sound is 0.250.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Frequency Change
The concept of frequency change is essential when discussing the Doppler Effect. Whether it's a train approaching or receding, the change in the sound's frequency is noticeably different when compared to a stationary observer. This difference in frequencies is a direct outcome of motion between the source and the observer.
  • Approaching Source: The frequency increases because the sound waves are compressed.
  • Receding Source: The frequency decreases as the sound waves are stretched out.
This change in frequency is described by the expressions for the detected frequencies:
- When moving toward the source, the frequency is given by: \[ f_{\text{app}}' = f \left( \frac{v + v_D}{v} \right) \]- When moving away from the source, the frequency is given by: \[ f_{\text{rec}}' = f \left( \frac{v - v_D}{v} \right) \]These formulas capture how the relative velocity influences the frequency observed. By understanding these changes, one can predict how sound behaves when there's movement involved.
Detector Speed
The speed of the detector, often denoted as \(v_D\), is crucial in the Doppler effect calculations. This speed refers to how fast the listening device (or an observer) is moving relative to the sound source. When the detector moves:
  • Toward the source: The detected frequency increases because the waves are encountered more frequently.
  • Away from the source: The frequency decreases as the waves are encountered less frequently.
In the context of the given problem, the detector's speed forms part of the equation that solves for the change in frequency:\[ \frac{f_{\text{app}}' - f_{\text{rec}}'}{f} = 0.500 \]By substituting the Doppler effect equations, we eventually find that the ratio of the detector's speed to the speed of sound, \( \frac{v_D}{v} \), is 0.250. Meaning, the detector moves at a quarter of the speed of sound for the given frequency change proportion.
Speed of Sound
The speed of sound, denoted as \(v\), significantly affects how sound waves travel in a medium. Typically, in air at 20°C, sound travels at approximately 343 meters per second.This speed determines how quickly the sound waves reach the observer or detector. In Doppler effect scenarios, such as the one described:
  • The formula for approaching frequencies depends heavily on this speed:
  •  \[ f_{\text{app}}' = f \left( \frac{v + v_D}{v} \right) \]
  • When receding, the speed is just as important:
  •  \[ f_{\text{rec}}' = f \left( \frac{v - v_D}{v} \right) \]
Sound speed in a medium like air doesn't change rapidly with small temperature variations, making our calculations reliable for everyday scenarios. This constancy allows us to make precise determinations about other variables, such as the detector speed compared to sound, as shown by finding the ratio \( \frac{v_D}{v} = 0.250 \).
Hence, knowing the speed of sound enables practical applications like measuring speed using Doppler radar.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At a certain point, two waves produce pressure variations given by \(\Delta p_{1}=\Delta p_{m} \sin \omega t\) and \(\Delta p_{2}=\Delta p_{m} \sin (\omega t-\phi) .\) At this point, what is the ratio \(\Delta p_{r} / \Delta p_{m},\) where \(\Delta p_{r}\) is the pressure amplitude of the resultant wave, if \(\phi\) is (a) \(0,\) (b) \(\pi / 2,\) (c) \(\pi / 3\), and (d) \(\pi / 4 ?\)

A stationary motion detector sends sound waves of frequency \(0.150 \mathrm{MHz}\) toward a truck approaching at a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). What is the frequency of the waves reflected back to the detector?

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.

An ambulance with a siren emitting a whine at \(1600 \mathrm{~Hz}\) over takes 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?

Two identical piano wires have a fundamental frequency of \(600 \mathrm{~Hz}\) when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of 6.0 beats/s when both wires oscillate simultaneously?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.