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A listener at rest (with respect to the air and the ground) hears a signal of frequency \(f_{1}\) from a source moving toward him with a velocity of \(15 \mathrm{~m} / \mathrm{s},\) due east. If the listener then moves toward the approaching source with a velocity of \(25 \mathrm{~m} / \mathrm{s},\) due west, he hears a frequency \(f_{2}\) that differs from \(f_{1}\) by \(37 \mathrm{~Hz}\). What is the frequency of the source? (Take the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\).)

Short Answer

Expert verified
The frequency of the source is 481 Hz.

Step by step solution

01

Define the Doppler Effect Equation for Sound

The Doppler Effect equation for the observed frequency when the source or listener is moving is given by \( f' = f \left( \frac{v + v_{o}}{v - v_{s}} \right) \) where \( f' \) is the observed frequency, \( f \) is the source frequency, \( v \) is the speed of sound, \( v_{o} \) is the speed of the observer, and \( v_{s} \) is the speed of the source. Note that when the listener and source move toward each other, their respective velocities are added to make the observed frequency higher.
02

Apply Doppler Effect When Source Moves Toward Listener at Rest

The listener is initially at rest and hears a frequency \( f_{1} \) from the source moving toward him with velocity \( v_{s} = 15 \) m/s. Using the Doppler effect, the observed frequency is \( f_{1} = f \left( \frac{340}{340 - 15} \right) \).
03

Apply Doppler Effect When Listener Moves Toward Source

Now, the listener moves toward the source with velocity \( v_{o} = 25 \) m/s while the source maintains its velocity. The new observed frequency is \( f_{2} = f \left( \frac{340 + 25}{340 - 15} \right) \).
04

Relate the Frequencies using Given Difference

From the problem, we know \( f_{2} - f_{1} = 37 \) Hz. Substituting the expressions for \( f_{1} \) and \( f_{2} \), we have:\[f \left( \frac{340 + 25}{340 - 15} \right) - f \left( \frac{340}{340 - 15} \right) = 37\]
05

Simplify the Equation and Solve for Source Frequency

Simplifying the equation from step 4, factor out \( f \):\[f \left[ \left( \frac{365}{325} \right) - \left( \frac{340}{325} \right) \right] = 37\]This simplifies to:\[f \left( \frac{25}{325} \right) = 37\]Solving for \( f \), we get:\[f = \frac{37 \times 325}{25} = 481\]

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Key Concepts

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

Frequency of Sound
The frequency of sound refers to how often the air pressure waves hit the listener's ear per second. It's measured in Hertz (Hz) and affects the pitch that we perceive. When discussing the Doppler effect, the concept of frequency becomes essential. The Doppler effect causes the observed frequency to change when either the source or the observer is moving relative to each other. The frequency emitted by the source remains constant, but how it is perceived depends on their relative velocities. For example, when a sound source approaches, the waves bunch up, leading to a higher frequency or pitch. Conversely, when the source moves away, the waves spread out, resulting in a lower frequency. Importantly, to calculate changes in observed frequency, we use the Doppler Effect equation, which considers both the speed of sound and the velocities of the source and listener. This addition or subtraction of velocities directly impacts the frequency heard by the observer.
Speed of Sound
The speed of sound is a fundamental constant for understanding how sound travels through air or any medium. For this exercise, the speed of sound in air is given as 340 m/s, which is a standard approximation at sea level and room temperature. The speed of sound plays a crucial role in the Doppler effect calculations. It represents the speed at which sound waves propagate through the medium. Variations in speed based on different media or conditions can alter how sound is perceived. Understanding this speed is important because it's the baseline value in the Doppler formula, against which the movements of listeners and sources are compared. In our problem, both the source and the observer's velocities are added or subtracted from this speed to determine the observed frequency. Thus, mastering this concept helps in grasping situations where sound frequency changes dynamically due to motion.
Wave Motion
Wave motion is how sound waves travel through the air from the source to the listener. These waves are mechanical waves, which require a medium, such as air, to propagate. Sound waves are longitudinal waves, meaning the air particles move parallel to the direction of wave travel. This is different from transverse waves, where the motion is perpendicular. Each wave consists of compressions and rarefactions. Compressions are regions where particles are closer together, while rarefactions are regions where they are farther apart. In the context of the Doppler effect, wave motion is essential because it explains how changes in frequency occur. The closer the sound source gets to the observer, the shorter the distance between successive waves, increasing the frequency. Conversely, as the source moves away, this distance increases, lowering the frequency. Understanding wave motion helps us visualize problems involving sound and aids in applying mathematical equations efficiently to real-world problems like those involving the Doppler effect.

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Most popular questions from this chapter

The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded together with a tuning fork that is oscillating accurately at concert A (440 Hz). What is the period of the violin string oscillation?

Pipe \(A\) has only one open end; pipe \(B\) is four times as long and has two open ends. Of the lowest 10 harmonic numbers \(n_{B}\) of pipe \(B,\) what are the (a) smallest, (b) second smallest, and (c) third smallest values at which a harmonic frequency of \(B\) matches one of the harmonic frequencies of \(A ?\)

A police car is chasing a speeding Porsche 911 . Assume that the Porsche's maximum speed is \(80.0 \mathrm{~m} / \mathrm{s}\) and the police car's is \(54.0 \mathrm{~m} / \mathrm{s}\). At the moment both cars reach their maximum speed, what frequency will the Porsche driver hear if the frequency of the police car's siren is \(440 \mathrm{~Hz}\) ? Take the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\).

A girl is sitting near the open window of a train that is moving at a velocity of \(10.00 \mathrm{~m} / \mathrm{s}\) to the east. The girl's uncle stands near the tracks and watches the train move away.The locomotive whistle emits sound at frequency \(500.0 \mathrm{~Hz}\). The air is still. (a) What frequency does the uncle hear? (b) What frequency does the girl hear? A wind begins to blow from the east at \(10.00 \mathrm{~m} / \mathrm{s} .\) (c) What frequency does the uncle now hear? (d) What frequency does the girl now hear?

Figure 17-48 shows an air filled, acoustic interferometer, used to demonstrate the interference of sound waves. Sound source \(S\) is an oscillating diaphragm; \(D\) is a sound detector, such as the ear or a microphone. Path \(S B D\) can be varied in length, but path \(S A D\) is fixed. At \(D,\) the sound wave coming along path \(S B D\) interferes with that coming along path \(S A D .\) In one demonstration, the sound intensity at \(D\) has a minimum value of 100 units at one position of the movable arm and continuously climbs to a maximum value of 900 units when that arm is shifted by \(1.65 \mathrm{~cm} .\) Find (a) the frequency of the sound emitted by the source and (b) the ratio of the amplitude at \(D\) of the \(S A D\) wave to that of the \(S B D\) wave. (c) How can it happen that these waves have different amplitudes, considering that they originate at the same source?

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