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Chapter 17: Problem 86

A sound source moves along an \(x\) axis, between detectors \(A\) and \(B .\) The wavelength of the sound detected at \(A\) is \(0.500\) that of the sound detected at \(B .\) What is the ratio \(v_{s} / v\) of the speed of the source to the speed of sound?

Short Answer

Expert verified
The ratio \( \frac{v_s}{v} \) is \( \frac{1}{3} \).

Step by step solution

01

Understanding the Problem

We are dealing with a classic Doppler effect scenario where a moving sound source emits waves that are detected differently at two points (A and B). The goal is to find the ratio of the source's speed \( v_s \) to the speed of sound \( v \). The wavelength detected at A is half that detected at B.
02

Applying Doppler Effect Formula

The Doppler effect equation for frequency when the source is moving towards the detector is \( f' = f \frac{v}{v - v_s} \) and away from the detector as \( f' = f \frac{v}{v + v_s} \). In terms of wavelength, these equations become \( \lambda' = \lambda \frac{v - v_s}{v} \) for moving to and \( \lambda' = \lambda \frac{v + v_s}{v} \) for moving away.
03

Relating Wavelengths to Speed Ratio

Given that the wavelength detected at A is \(0.500\) that at B, we identify \( \lambda_A = \frac{v - v_s}{v} \lambda \) and \( \lambda_B = \frac{v + v_s}{v} \lambda \). Thus, \( \frac{v - v_s}{v} / \frac{v + v_s}{v} = 0.500 \).
04

Solving the Equation

We simplify the equation found in the previous step: \( \frac{v - v_s}{v + v_s} = 0.500 \). By doing cross-multiplication, we arrive at: \( v - v_s = 0.500(v + v_s) \). Simplifying, \( v - v_s = 0.500v + 0.500v_s \).
05

Isolating the Source Speed

Rearrange the equation to isolate \( v_s \): \( v - 0.500v = 1.500v_s \). Simplifying further gives \( 0.500v = 1.500v_s \). Thus, \( v_s = \frac{0.500}{1.500} v \).
06

Finding the Final Ratio

Simplifying the fraction gives \( v_s = \frac{1}{3}v \). Hence, the ratio \( \frac{v_s}{v} = \frac{1}{3} \).

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Key Concepts

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

Understanding Wave Frequency
When we talk about wave frequency in the context of the Doppler effect, it's crucial to grasp that frequency refers to the number of wave cycles that pass a given point in one second. This is usually expressed in Hertz (Hz). Wave frequency changes depending on the motion of the source relative to the observer. If the source of the wave moves toward the observer, the frequency increases, meaning there are more waves per second. Conversely, if the source moves away, the frequency decreases.

This phenomenon explains why we hear a change in pitch as a vehicle approaches and then moves away from us. Understanding how frequency is affected by motion is key to solving physics problems involving the Doppler effect.

In the given problem, two detectors measure sound waves at different frequencies due to the source's motion. Recognizing this change is essential for calculating the relative speeds of the sound source and the sound itself.
Exploring Wave Speed
Wave speed is another fundamental concept. It refers to the speed at which a wave travels through a medium. For sound waves, this speed depends on the medium's properties such as its temperature, density, and elasticity.

The Doppler effect comes into play when the emitter or the detector moves. The relative motion causes changes in the observed frequencies and, therefore, the wavelengths detected by different observers. The constant wave speed remains the characteristic speed of sound in the medium, unaffected by the observer's movement.

In our exercise, we're tasked with finding the ratio of the source speed to this constant wave speed of sound. Using the Doppler effect formula allows us to observe how the change in wavelength detected by different observers can be related back to the movement of the source, helping us determine this ratio effectively.
Physics Problem-Solving Strategies
Physics problem-solving often requires a structured approach. For Doppler effect problems like the one discussed, it's essential to understand the problem context completely before applying formulas.

Here's a quick guide to tackle such problems:
  • Understand what the problem is asking. Identify known values and unknowns.
  • Apply relevant physics equations to bridge the known values to the unknowns.
  • Simplify equations logically, performing consistent algebraic manipulations to isolate desired variables.
In our exercise, the process involved understanding how the motion of the sound source affected the detected wavelengths, applying the Doppler effect formula for wavelength, and solving algebraically for the speed ratio.

Overall, the key to mastering physics problems is to break them down into smaller problems, methodically solve each one, and assemble the results for the final solution. This methodical approach fosters confidence and efficiency in problem-solving.

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Most popular questions from this chapter

Two trains are traveling toward each other at \(30.5 \mathrm{~m} / \mathrm{s}\) relative to the ground. One train is blowing a whistle at \(500 \mathrm{~Hz}\). (a) What frequency is heard on the other train in still air? (b) What frequency is heard on the other train if the wind is blowing at \(30.5 \mathrm{~m} / \mathrm{s}\) toward the whistle and away from the listener? (c) What frequency is heard if the wind direction is reversed?

Hot chocolate effect. Tap a metal spoon inside a mug of water and note the frequency \(f_{i}\) you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value \(f_{s}\) because the tiny air bubbles released by the powder change the water's bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don't appreciably change the water's density or volume or the sound's wavelength.

Two loud speakers are located \(3.35 \mathrm{~m}\) apart on an outdoor stage. A listener is \(18.3 \mathrm{~m}\) from one and \(19.5 \mathrm{~m}\) from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range \((20 \mathrm{~Hz}\) to \(20 \mathrm{kHz}) .\) (a) What is the lowest frequency \(f_{\min , 1}\) that gives minimum signal (destructive interference) at the listener's location? By what number must \(f_{\min , 1}\) be multiplied to get (b) the second lowest frequency \(f_{\min 2}\) that gives minimum signal and (c) the third lowest frequency \(f_{\min , 3}\) that gives minimum signal? (d) What is the lowest frequency \(f_{\max , 1}\) that gives maximum signal (constructive interference) at the listener's location? By what number must \(f_{\max , 1}\) be multiplied to get (e) the second lowest frequency \(f_{\max , 2}\) that gives maximum signal and (f) the third lowest frequency \(f_{\max , 3}\) that gives maximum signal?

A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is \(39000 \mathrm{~Hz}\). During one fast swoop directly toward a flat wall surface, the bat is moving at \(0.025\) times the speed of sound in air. What frequency does the bat hear reflected off the wall?

A sound source \(A\) and a reflecting surface \(B\) move directly toward each other. Relative to the air, the speed of source \(A\) is \(29.9 \mathrm{~m} / \mathrm{s}\), the speed of surface \(B\) is \(65.8 \mathrm{~m} / \mathrm{s}\), and the speed of sound is \(329 \mathrm{~m} / \mathrm{s}\). The source emits waves at frequency \(1200 \mathrm{~Hz}\) as measured in the source frame. In the reflector frame, what are the (a) frequency and (b) wavelength of the arriving sound waves? In the source frame, what are the (c) frequency and (d) wavelength of the sound waves reflected back to the source?
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