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A source emitting sound at frequency \(4000 \mathrm{~Hz}\), is moving along the \(Y\) -axis with a speed of \(22 \mathrm{~m} \mathrm{~s}^{-1} .\) A listener is situated on the ground at the position \((660 \mathrm{~m}, 0)\). Find the frequency of the sound received by the listener at the instant the source crosses the origin. Speed of sound in air \(=330 \mathrm{~m} \mathrm{~s}^{-1}\).

Short Answer

Expert verified
4000 Hz, as there's no Doppler shift when the source crosses perpendicularly.

Step by step solution

01

Understand the Doppler Effect

The Doppler Effect describes the change in frequency (or wavelength) of a wave in relation to an observer moving relative to the source of the wave. Depending on whether the source is moving towards or away from the observer, the frequency perceived by the observer will increase or decrease.
02

Identify Variables

From the problem, we have: a sound source frequency, \(f_s = 4000 \, \text{Hz}\), source speed \(v_s = 22 \, \text{m/s}\), speed of sound \(v = 330 \, \text{m/s}\). The listener is stationary, so their speed \(v_l = 0 \, \text{m/s}\).
03

Calculate the Effective Velocity

When the source crosses the origin, its motion is perpendicular to the line connecting the source and the listener. To account for this, calculate the component of the source's velocity towards the listener, which at crossing is zero since the source is moving vertically and not directly towards or away from the listener.
04

Apply the Doppler Effect Formula

The formula for the observed frequency \(f\) when the source is moving perpendicular to the line joining the observer is: \[ f = f_s \frac{v}{v} \] as there's no component of velocity along the line of the listener. Hence, \(f = f_s = 4000 \, \text{Hz}\).
05

Interpret the Formula

Since the source is moving perpendicularly right at the instant it crosses the origin, the frequency perceived by the listener does not change from the source frequency, as there is no relative motion towards or away from the listener.

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Key Concepts

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

Sound Frequency
Sound frequency refers to the number of vibrations or oscillations of sound waves that occur in one second. It is measured in Hertz (Hz). Higher frequencies are perceived as higher pitches, while lower frequencies sound deeper. For the given exercise, the source emits sound at a frequency of 4000 Hz.
This frequency represents how many times a sound wave oscillates every second. Many musical notes and sounds we hear around us are categorized by their frequencies, which determine their perceived pitches.
In this particular exercise, understanding the emitted sound frequency is crucial, as it directly affects how the Doppler Effect influences what is heard by an observer. Recognizing the change or constancy in perceived sound frequency provides insights into movement characteristics of the sound source.
Source Velocity
Source velocity refers to the speed of the sound source as it moves through space. In this exercise, the source moves at 22 m/s along the Y-axis.
The velocity of a source affects how its sound waves compress or spread out, influencing the frequency perceived by an observer. When a source approaches an observer, sound waves get compressed, raising the frequency and pitch; if the source moves away, the sound expands, lowering these attributes.
In this scenario, the source's velocity is crucial when applying the Doppler Effect. However, at the moment when the source is directly perpendicular to the observer (crossing the origin), its velocity component towards the observer becomes zero, resulting in no frequency shift at that instant.
Wave Perception
Wave perception pertains to how sound waves are interpreted by an observer, which can be influenced by multiple factors like the relative motion of source and listener, and environmental conditions.
The Doppler Effect plays a key role in wave perception, where perceived sound frequency can change due to the movement of either the source or the listener. This effect mirrors real-world experiences like hearing a siren change pitch as it moves past you.
In the exercise, since the source crosses the origin perpendicularly, wave perception by the listener does not change. The perceived frequency is the same as the emitted frequency, 4000 Hz, as the motion has no impact along the listener's direction.
Speed of Sound
Speed of sound in air is a constant value, commonly taken as 330 m/s for calculations under typical conditions. This speed determines how quickly sound travels from the source to an observer, influencing how soon a sound is perceived.
Factors such as air temperature and pressure can slightly alter this speed, but in many calculations, like this exercise, it's considered constant to simplify the process.
In the given problem, the speed of sound is essential in applying the Doppler Effect formula correctly. Since there’s no velocity component change towards the listener at the instance of crossing the origin, the sound frequency remains unchanged, signifying the steady and predictable influence of sound speed on frequency calculations.

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Most popular questions from this chapter

Two coherent narrow slits emitting sound of wavelength \(\lambda\) in the same phase are placed parallel to each other at a small separation of \(2 \lambda\). The sound is detected by moving a detector on the screen \(\Sigma\) at a distance \(D(\gg>)\) from the slit \(S_{1}\) as shown in figure (16-E6). Find the distance \(x\) such that the intensity at \(P\) is equal to the intensity at \(O\).

In Quincke's experiment, the sound intensity has a minimum value \(I\) at a particular position. As the sliding tube is pulled out by a distance of \(16: 5 \mathrm{~mm}\), the intensity increases to a maximum of \(9 I .\) Take the speed of sound in air to be \(330 \mathrm{~m} \mathrm{~s}^{-1}\). (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.

A boy riding on a bicycle going at \(12 \mathrm{~km} \mathrm{~h}^{-1}\) towards a vertical wall whistles at his dog on the ground. If the frequency of the whistle is \(1600 \mathrm{~Hz}\) and the speed of sound in air is \(330 \mathrm{~m} \mathrm{~s}^{-1}\), find (a) the frequency of the whistle as received by the wall (b) the frequency of the reflected whistle as received by the boy.

The separation between a node and the next antinode in a vibrating air column is \(25 \mathrm{~cm}\). If the speed of sound in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\), find the frequency of vibration of the air column.

A source of sound operates at \(2 \cdot 0 \mathrm{kHz}, 20 \mathrm{~W}\) emitting sound uniformly in all directions. The speed of sound in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\) and the density of air is \(1 \cdot 2 \mathrm{~kg} \mathrm{~m}^{-3}\). (a) What is the intensity at a distance of \(6 \cdot 0 \mathrm{~m}\) from the source? (b) What will be the pressure amplitude at this point? (c) What will be the displacement amplitude at this point?

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