/*! 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 16 An automobile moving at \(30.0 \... [FREE SOLUTION] | 91Ó°ÊÓ

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An automobile moving at \(30.0 \mathrm{~m} / \mathrm{s}\) is approaching a factory whistle that has a frequency of \(500 \mathrm{~Hz}\). (a) If the speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\), what is the apparent frequency of the whistle as heard by the driver? (b) Repeat for the case of the car leaving the factory at the same speed. This is a Doppler shift problem. Draw an arrow from observer to source; this is the positive direction. Here in part \((a)\) the observer is moving in the positive direction, and \(u_{s}=0 .\) Hence, use \(+u_{o}\) and So (a) \(f_{f}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=544 \mathrm{~Hz}\) With the car leaving in the negative direction use \(-\mathrm{u}_{o}\) and (b) \(f_{n}=f_{0} \frac{v \pm v_{o}}{v \mp v_{3}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}-30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=456 \mathrm{~Hz}\)

Short Answer

Expert verified
(a) 544 Hz when approaching, (b) 456 Hz when receding.

Step by step solution

01

Identify the Given Values

We are provided with the following values: the speed of the car \(v_o = 30.0\, \mathrm{m/s}\), the frequency of the whistle \(f_s = 500\, \mathrm{Hz}\), and the speed of sound in air \(v = 340\, \mathrm{m/s}\).
02

Understand the Doppler Effect Formula

In the Doppler effect, the apparent frequency \(f'\) heard by an observer moving towards the source is given by: \[ f' = f_s \left( \frac{v + v_o}{v - v_s} \right) \] where \(v_s\) is the speed of the source. Since the source is stationary, \(v_s = 0\).
03

Calculate Apparent Frequency as the Car Approaches (a)

Substitute the values in the formula considering \(+v_o\) because the car is moving towards the source: \[ f' = 500\, \mathrm{Hz} \left( \frac{340\, \mathrm{m/s} + 30\, \mathrm{m/s}}{340\, \mathrm{m/s} - 0} \right) \] This simplifies to: \[ f' = 500 \times \frac{370}{340} \approx 544\, \mathrm{Hz} \] So, the apparent frequency is \(544\, \mathrm{Hz}\) when approaching.
04

Calculate Apparent Frequency as the Car Recedes (b)

This time, because the car is moving away, we use \(-v_o\): \[ f' = 500\, \mathrm{Hz} \left( \frac{340\, \mathrm{m/s} - 30\, \mathrm{m/s}}{340\, \mathrm{m/s} - 0} \right) \] Simplifying, we get: \[ f' = 500 \times \frac{310}{340} \approx 456\, \mathrm{Hz} \] Therefore, the apparent frequency is \(456\, \mathrm{Hz}\) when receding.

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Key Concepts

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

Apparent Frequency
In the context of the Doppler Effect, the apparent frequency is the frequency perceived by an observer who is moving relative to the source of the sound. This frequency can differ from the actual frequency emitted by the source. The apparent frequency is higher when the observer moves towards the source and lower when moving away. This change occurs because the motion alters the wavelength reaching the observer, without changing the actual frequency emitted by the source.
The formula used to determine the apparent frequency is:
  • When the observer approaches the source: \[ f' = f_s \left( \frac{v + v_o}{v - v_s} \right) \]
  • When the observer moves away: \[ f' = f_s \left( \frac{v - v_o}{v - v_s} \right) \]
Understanding apparent frequency is crucial in situations where an observer or the source is in motion, such as in vehicles or when analyzing waves in various mediums.
Speed of Sound
The speed of sound is a critical factor in calculating the Doppler Effect. In this particular exercise, the speed of sound in air is given as \(340 \text{ m/s}\). This is the rate at which sound waves travel through the air and it influences how quickly the sound reaches the observer from the source.
The speed of sound can vary based on several factors:
  • Temperature: Sound travels faster in warmer air.
  • Medium: Sound speed changes with different materials, like water or metals.
  • Humidity and pressure also have an impact, but usually less significant.
In problems involving sound and motion, using the correct speed of sound is vital to accurately determine the apparent frequency using the Doppler Effect formula.
Observer Motion
Observer motion is a key component in understanding how the Doppler Effect changes the perception of sound frequency. In this exercise, the observer is in an automobile moving at \(30.0 \text{ m/s}\).
Whether the observer is approaching or receding from the sound source influences how the sound frequency is perceived:
  • Approaching the source: The motion towards the source compresses the sound waves, making the frequency higher.
  • Receding from the source: The motion away from the source stretches the waves, resulting in a lower frequency.
It’s crucial to correctly identify the direction of motion since it determines if \(+v_o\) or \(-v_o\) is used in the Doppler Effect equation.
Source Frequency
The source frequency is the actual frequency at which the sound waves are emitted by the source. In our scenario, the factory whistle produces sound at a frequency of \(500 \text{ Hz}\).
This frequency remains constant irrespective of other factors like motion or medium changes. It serves as the baseline for calculating the apparent frequency when the Doppler Effect is in play.
It's vital to note:
  • Source frequency does not change due to the motion of the observer or the medium; it is an attribute of the sound source itself.
  • It provides a reference point against which the perceived changes due to motion are measured.
Knowing the source frequency allows for the use of the Doppler Effect formula to determine how the frequency will be perceived under different conditions of motion and speed.

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Most popular questions from this chapter

Three seconds after a gun is fired, the person who fired the gun hears an echo. How far away was the surface that reflected the sound of the shot? Use \(340 \mathrm{~m} / \mathrm{s}\) for the speed of sound.

The two sound sources in vibrate in-phase. A loud sound is heard at \(P\) when \(L_{1}=L_{2}\). As \(L_{1}\) is slowly increased, the weakest sound is heard when \(L_{1}-L_{2}\) has the values \(20.0 \mathrm{~cm}, 60.0 \mathrm{~cm}\), and \(100 \mathrm{~cm}\). What is the frequency of the sound source if the speed of sound is \(340 \mathrm{~m} / \mathrm{s} ?\) The waves coming down directly from the fork toward the guy must be a little longer (more spaced) than the waves going up from the fork and back down from the wall, which have the same spacings. The weakest sound will be heard at \(P\) when a crest from \(S_{1}\) and a trough from \(S_{2}\) reach there at the same time. This will happen if \(L_{1}\) \(-L_{2}\) is \(\frac{1}{2} \lambda\), or \(\lambda+\frac{1}{2} \lambda\), or \(2 \lambda+\frac{1}{2} \lambda\), and so on. Hence, the increase in \(L_{1}\) between weakest sounds is \(\lambda\), and from the data we see that \(\lambda=\) \(0.400 \mathrm{~m}\). Then, from \(\lambda=\mathrm{v} / \mathrm{f}\), $$ f=\frac{v}{\lambda}=\frac{340 \mathrm{~m} / \mathrm{s}}{0.400 \mathrm{~m}}=850 \mathrm{~Hz} $$

A shell fired at a target \(800 \mathrm{~m}\) away was heard by someone standing near the gun to strike the target \(5.0 \mathrm{~s}\) after leaving the gun. Compute the average horizontal velocity of the shell. The air temperature is \(20^{\circ} \mathrm{C}\).

Suppose the intensity of sound increases by a multiplicative factor of \(10.0\), going from say \(1.0 \mathrm{~W} / \mathrm{m}^{2}\) to \(10 \mathrm{~W} / \mathrm{m}^{2}\) to \(100 \mathrm{~W} / \mathrm{m}^{2}\) and so on. By how much is the intensity level increased each time?

Two cars are heading straight at each other with the same speed. The horn of one \((f=3.0 \mathrm{kHz})\) is blowing, and is heard to have a frequency of \(3.4 \mathrm{kHz}\) by the people in the other car. Find the spee at which each car is moving if the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).

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