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Chapter 17: Problem 65

A \(2000 \mathrm{~Hz}\) siren and a civil defense official are both at rest with respect to the ground. What frequency does the official hear if the wind is blowing at \(12 \mathrm{~m} / \mathrm{s}\) (a) from source to official and (b) from official to source?

Short Answer

Expert verified
(a) 1932.4 Hz; (b) 2072.5 Hz.

Step by step solution

01

Understanding the scenario and given data

We need to determine the frequency heard by the official considering the effect of wind. The scenarios are: (a) wind blowing from source to official, and (b) wind blowing from official to source. The given data are: source frequency \( f_s = 2000 \text{ Hz} \), wind speed \( v_ ext{wind} = 12 \text{ m/s} \), and the speed of sound \( v = 343 \text{ m/s} \) (standard speed of sound in air at 20掳C).
02

Choice of formula and variable identification

We use the Doppler effect formula to determine the frequency heard by an observer in the presence of wind: \[ f' = f_s \times \frac{v + v_o}{v + v_s} \]where \( f' \) is the observed frequency, \( v_o \) is the speed of the observer (0 m/s since the official is at rest), and \( v_s \) is the speed of the source (0 m/s since the siren is at rest). The effect of wind is incorporated by altering the speed of sound.
03

Calculating frequency for wind from source to official

For wind blowing from the source to the official, the wind aids the propagation of sound, thus effectively increasing the speed of sound relative to the observer. We modify the speed of sound as \( v' = v + 12 \text{ m/s} \). Plug in the values:\[ f' = 2000 \times \frac{343 + 0}{343 + 0 + 12} = 2000 \times \frac{343}{355} \approx 1932.39 \text{ Hz} \]The frequency heard by the official is approximately 1932.4 Hz.
04

Calculating frequency for wind from official to source

For wind blowing from the official to the source, the wind opposes the propagation of sound, effectively decreasing the speed of sound relative to the observer. We modify the speed of sound as \( v' = v - 12 \text{ m/s} \):\[ f' = 2000 \times \frac{343}{343 - 12} = 2000 \times \frac{343}{331} \approx 2072.51 \text{ Hz} \]The frequency heard by the official is approximately 2072.5 Hz.

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Key Concepts

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

Frequency Calculation
To understand the Doppler Effect, we begin with calculating the observed frequency using given conditions. Here, the original frequency of a siren is 2000 Hz, and we need to consider how wind affects this frequency. First, let's use the Doppler Effect formula, where \[ f' = f_s \times \frac{v + v_o}{v + v_s} \]is crucial. In this formula:
  • \( f' \) is the observed frequency.
  • \( f_s \) is the source frequency, given as 2000 Hz.
  • \( v \) is the speed of sound, typically 343 m/s in air at 20掳C.
  • \( v_o \) is the observer's speed, which is 0 m/s since the official is stationary.
  • \( v_s \) is the source's speed, also 0 m/s given the siren is stationary.
The Doppler effect essentially changes the frequency due to relative motion, and here, we modify it for wind conditions, treating the wind as altering the speed of sound. Understanding these modifications to the formula is key to solving such problems.
Wind Speed
Wind speed plays a crucial role in sound propagation by either aiding or opposing the travel of sound waves. When evaluating the scenario of sound traveling from a source to an official, the wind's direction is vital.Let's consider two different cases:
  • Wind from Source to Official: Here, the wind aids the sound wave, essentially increasing the sound speed as experienced by the observer. This is expressed through modified sound speed, given by \( v' = v + 12 \text{ m/s} \). Hence, the wind makes it easier for the sound to reach the listener.
  • Wind from Official to Source: In this case, the wind opposes the sound wave's journey, effectively reducing the speed of sound as experienced by the observer. The modified sound speed is \( v' = v - 12 \text{ m/s} \), which means sound travels slower relative to the observer due to wind's counteraction.
Thus, wind impact results in different observed frequencies in each scenario. Grasping these effects clarifies the differences in sound perception with respect to wind direction.
Sound Propagation
Sound propagation involves understanding how sound waves travel from a source to a listener. In the presence of environmental factors like wind, sound can be affected significantly. The basic speed of sound is influenced by temperature, but environmental effects like wind also alter it considerably.
  • Without Wind Influence: The sound travels at a standard speed, calculated as approximately 343 m/s at around 20掳C.
  • With Wind Influence: When wind is present, the effective speed is adjusted by the wind speed. If the wind goes from the source to the listener, sound propagation is aided, raising the effective speed. Conversely, wind direction from listener to source subtracts from the speed, causing a delay.
The result is a different frequency heard by the listener, even though the source frequency remains constant. This understanding of sound propagation, complemented by wind variation, informs us about how sound travels through the environment and is perceived differently.

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Most popular questions from this chapter

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?

One of the harmonic frequencies of tube \(A\) with two open ends is \(325 \mathrm{~Hz}\). The next-highest harmonic frequency is \(390 \mathrm{~Hz}\). (a) What harmonic frequency is next highest after the harmonic frequency \(195 \mathrm{~Hz}\) ? (b) What is the number of this next-highest harmonic? One of the harmonic frequencies of tube \(B\) with only one open end is \(1080 \mathrm{~Hz}\). The next-highest harmonic frequency is 1320 \(\mathrm{Hz}\). (c) What harmonic frequency is next highest after the harmonic frequency \(600 \mathrm{~Hz} ?\) (d) What is the number of this next-highest harmonic?

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?

Find the ratios (greater to smaller) of the (a) intensities, (b) pressure amplitudes, and (c) particle displacement amplitudes for two sounds whose sound levels differ by \(37 \mathrm{~dB}\).

A sinusoidal sound wave moves at \(343 \mathrm{~m} / \mathrm{s}\) through air in the positive direction of an \(x\) axis. At one instant, air molecule \(A\) is at its maximum displacement in the negative direction of the axis while air molecule \(B\) is at its equilibrium position. The separation between those molecules is \(15.0 \mathrm{~cm}\), and the molecules between \(A\) and \(B\) have intermediate displacements in the negative direction of the axis. (a) What is the frequency of the sound wave? In a similar arrangement, for a different sinusoidal sound wave, air molecule \(C\) is at its maximum displacement in the positive direction while molecule \(D\) is at its maximum displacement in the negative direction. The separation between the molecules is again \(15.0 \mathrm{~cm}\), and the molecules between \(C\) and \(D\) have intermediate displacements. (b) What is the frequency of the sound wave?
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