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Two train whistles, \(A\) and \(B\), each have a frequency of \(392 \mathrm{~Hz}\). \(A\) is stationary and \(B\) is moving toward the right (away from \(A\) ) at a speed of \(35.0 \mathrm{~m} / \mathrm{s}\). A listener is between the two whistles and is moving toward the right with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). (See Figure \(12.43 .)\) (a) What is the frequency from \(A\) as heard by the listener? (b) What is the frequency from \(B\) as heard by the listener? (c) What is the beat frequency detected by the listener?

Short Answer

Expert verified
(a) 409.14 Hz, (b) 372.57 Hz, (c) 36.57 Hz

Step by step solution

01

Determine the frequency heard from whistle A

Whistle A is stationary. According to the Doppler Effect formula for a sound source with no motion, the frequency heard by the listener, moving towards the stationary source, is given by: \[ f' = \left( \frac{v + v_L}{v} \right) f \]Here, \(v = 343 \text{ m/s}\) is the speed of sound in air, \(v_L = 15.0 \text{ m/s}\) is the velocity of the listener moving towards the source, and \(f = 392 \text{ Hz}\) is the original frequency of the whistle A. Substituting the values, we get:\[ f' = \left(\frac{343 + 15}{343}\right) \times 392 \approx 409.14 \text{ Hz} \]
02

Determine the frequency heard from whistle B

Whistle B is moving away from the listener who is moving towards it. The Doppler Effect formula to find the frequency heard from a moving source is:\[ f' = \left( \frac{v + v_L}{v + v_s} \right) f \]Where \(v_s = 35.0 \text{ m/s}\) is the speed of source B moving away. Plugging in the values, we have:\[ f' = \left(\frac{343 + 15}{343 + 35}\right) \times 392 \approx 372.57 \text{ Hz} \]
03

Calculate the beat frequency

The beat frequency is the absolute difference between the two frequencies heard by the listener. Using the frequencies calculated in Steps 1 and 2:\[ f_{\text{beat}} = |f'_A - f'_B| \approx |409.14 - 372.57| \approx 36.57 \text{ Hz} \]

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Key Concepts

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

Beat Frequency
Beat frequency is an interesting phenomenon that occurs when two sound waves of slightly different frequencies are heard simultaneously. It results in a pulsing variation in loudness called beats. These beats can be heard as periodic rises and falls in volume, much like a beating drum.

To understand beat frequency better, consider two sound sources emitting sounds of frequencies close to each other, like in the train whistle example in the exercise. The beat frequency, denoted as \( f_{\text{beat}} \), is calculated using the formula:
  • \( f_{\text{beat}} = |f'_A - f'_B| \)
This formula shows that the beat frequency is simply the absolute difference between the two perceived frequencies. In our case, it reflects the variation in sound as detected by a listener between two frequencies altered by the Doppler Effect.
The listener hears these beats as a result of constructive and destructive interference between the two sound waves, which leads to these fluctuations in intensity.
Sound Wave Frequency
Sound wave frequency is an essential concept in understanding waves and their behaviors. Frequency, denoted by \( f \), is the number of complete wave cycles that pass a point in one second. It is measured in Hertz (Hz).

Higher frequency sound waves correspond to higher-pitched sounds, while lower frequencies are associated with deeper sounds. In the exercise, both train whistles are originally producing sound waves at a frequency of 392 Hz. However, due to the relative movements of the sources and the listener, these frequencies change.
Using the Doppler Effect, we calculate the changes in frequency when either the source or listener, or both, are moving. This effect accounts for the listener hearing one frequency slightly higher from whistle A and another slightly lower from whistle B.
Speed of Sound
The speed of sound, commonly denoted as \( v \), is the rate at which sound waves travel through a medium. For air, which is a common medium, the speed of sound is approximately 343 m/s at room temperature.

This speed can vary based on factors like temperature, humidity, and air pressure. In our exercise, we use the speed of sound in our Doppler Effect calculations to determine how the motion of the source and listener affects the sound wave frequency as heard by the listener.
Sound travels faster in warmer conditions as the air molecules move more rapidly, aiding in the sound wave's propagation. This means that when solving problems like those in the exercise, it is crucial to use the correct speed of sound based on environmental conditions.
Velocity of Source and Listener
The velocity of the source and listener are key variables in understanding the Doppler Effect. **Velocity** refers to the speed and direction of motion of an object.
  • In the exercise, the source whistle B has a velocity of 35 m/s moving away from the stationary whistle A.
  • The listener, positioned between these two sources, moves at 15 m/s in the same direction as B.
Understanding their velocities is crucial when using the Doppler Effect formulas. The listener moving towards the source or away from it affects the frequency of the sound perceived.
For our calculations:
  • When the listener moves towards the sound source (like whistle A), the frequency heard increases.
  • When the listener moves in the same direction as a retreating source (like whistle B), the frequency heard decreases.
Hence, by incorporating their velocities into the Doppler Effect formula, we can account for changes in perceived sound frequency.

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Most popular questions from this chapter

A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pitch. The first-chair player's note is exactly \(440 \mathrm{~Hz}\), and 2.8 beats per second are heard. What are the two possible frequencies of the other player's note?

With what tension must a rope with length \(2.50 \mathrm{~m}\) and mass \(0.120 \mathrm{~kg}\) be stretched for transverse waves of frequency \(40.0 \mathrm{~Hz}\) to have a wavelength of \(0.750 \mathrm{~m} ?\)

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is \(10.0 \%\) higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

\(\mathrm{A} 75.0 \mathrm{~cm}\) wire of mass \(5.625 \mathrm{~g}\) is tied at both ends and adjusted to a tension of \(35.0 \mathrm{~N}\). When it is vibrating in its second overtone, find (a) the frequency and wavelength at which it is vibrating and (b) the frequency and wavelength of the sound waves it is producing.

One of the \(63.5-\mathrm{cm}\) -long strings of an ordinary guitar is tuned to produce the note \(\mathrm{B}_{3}\) (frequency \(245 \mathrm{~Hz}\) ) when vibrating in its fundamental mode. (a) Find the speed of transverse waves on this string. (b) If the tension in this string is increased by \(1.0 \%,\) what will be the new fundamental frequency of the string? (c) If the speed of sound in the surrounding air is \(344 \mathrm{~m} / \mathrm{s},\) find the frequency and wavelength of the sound wave produced in the air by the vibration of the \(\mathrm{B}_{3}\) string. How do these compare to the frequency and wavelength of the standing wave on the string?

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