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Chapter 17: Problem 58

A sound source \(A\) and a reflecting surface \(B\) move directly toward each other. Relative to the air, the speed of source \(A\) is \(29.9\) \(\mathrm{m} / \mathrm{s}\), the speed of surface \(B\) is \(65.8 \mathrm{~m} / \mathrm{s}\), and the speed of sound is \(329 \mathrm{~m} / \mathrm{s}\). The source emits waves at frequency \(1200 \mathrm{~Hz}\) as measured in the source frame. In the reflector frame, what are the (a) frequency and (b) wavelength of the arriving sound waves? In the source frame, what are the (c) frequency and (d) wavelength of the sound waves reflected back to the source?

Short Answer

Expert verified
(a) 1435 Hz, (b) 0.229 m, (c) 1738 Hz, (d) 0.189 m

Step by step solution

01

Understanding the situation and parameters

We have a sound source \(A\) moving at \(29.9\, \text{m/s}\) towards the reflector \(B\), which itself moves towards \(A\) at \(65.8\,\text{m/s}\). The speed of sound is \(329\,\text{m/s}\). The source emits a frequency of \(1200\,\text{Hz}\). We'll calculate how the movement affects frequency and wavelength as perceived in different reference frames.
02

Calculating frequency in reflector frame (a)

Using the Doppler effect formula for when both the source and the observer are moving towards each other:\[ f' = \frac{v + v_o}{v - v_s} \cdot f \]Where:- \(v = 329\,\text{m/s}\) (speed of sound)- \(v_o = 65.8\,\text{m/s}\) (speed of reflector \(B\))- \(v_s = 29.9\,\text{m/s}\) (speed of source \(A\))- \(f = 1200\,\text{Hz}\)Substitute the values:\[ f' = \frac{329 + 65.8}{329 - 29.9} \times 1200 \approx 1435\,\text{Hz} \]
03

Calculating wavelength in reflector frame (b)

The wavelength \( \lambda' \) in the reflector frame can be found using the relationship:\[ \lambda' = \frac{v}{f'} \]Substitute \(v = 329\,\text{m/s}\) and previously calculated \(f' \approx 1435\,\text{Hz}\):\[ \lambda' = \frac{329}{1435} \approx 0.229\,\text{m} \]
04

Calculating frequency of reflected sound in source frame (c)

After reflection, \(B\) becomes the source and \(A\) becomes the observer. Use the Doppler formula again, now with positions reversed:\[ f'' = \frac{v + v_s}{v - v_o} \cdot f' \]Substitute previously calculated \(f'\):\[ f'' = \frac{329 + 29.9}{329 - 65.8} \times 1435 \approx 1738\,\text{Hz} \]
05

Calculating wavelength of reflected sound in source frame (d)

Use the relationship between wavelength and frequency again:\[ \lambda'' = \frac{v}{f''} \]Substitute \(f'' \approx 1738\,\text{Hz}\):\[ \lambda'' = \frac{329}{1738} \approx 0.189\,\text{m} \]

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Key Concepts

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

Sound waves
Sound waves are a type of mechanical wave that travels through a medium, such as air, water, or solid materials. They are created by vibrating objects, which cause variations in pressure that move through the medium. This type of wave needs a medium to travel and cannot exist in a vacuum, unlike electromagnetic waves.
The speed at which sound waves travel depends on the medium. In air at room temperature, this speed is approximately 343 meters per second, but it can vary depending on conditions such as temperature and pressure. In our exercise, the given speed of sound in air is specified as 329 meters per second.
It's important to understand that the properties of sound waves, such as frequency and wavelength, can change depending on the movement of the source or observer. This principle is precisely what the Doppler Effect explains, allowing us to calculate changes in perceived sound frequency and wavelength due to motion.
Frequency calculation
Frequency refers to how often something occurs in a given period. For sound waves, it is the number of vibrations or cycles per second, measured in hertz (Hz). Frequency determines the pitch of the sound; higher frequencies correspond to higher pitches.
When the sound source and the observer are in motion relative to each other, like in our example with the moving source and reflector, the frequency of the observed sound can differ from the frequency at which it was emitted. This change is what the Doppler Effect describes.
To calculate the impact of movement on frequency, we use the Doppler Effect formula:
  • For the arriving sound waves, the formula is: \[ f' = \frac{v + v_o}{v - v_s} \cdot f \] where:
    • \(v\) is the speed of sound, \(v_o\) is the speed of the observer, \(v_s\) is the speed of the source, and \(f\) is the original frequency.
  • After reflection, use: \[ f'' = \frac{v + v_s}{v - v_o} \cdot f' \] where the positions of the source and observer are swapped.
In our example, using these formulas, the frequency changes first to approximately 1435 Hz in the reflector frame and then to about 1738 Hz in the source frame.
Wavelength calculation
Wavelength is the distance between consecutive peaks of a wave. In sound waves, it is how far the wave travels during one complete cycle of the vibration that generates it. Wavelength and frequency are closely linked through the speed of sound.
The formula to relate these is:\[ \lambda = \frac{v}{f} \]where \(\lambda\) is the wavelength, \(v\) is the speed of sound, and \(f\) is the frequency.
When calculating wavelengths in different frames, we need to adjust for changes in frequency due to the Doppler Effect. In our exercise, we first found that the wavelength in the reflector frame is about 0.229 meters using the frequency of 1435 Hz. Then, using the frequency of 1738 Hz, the wavelength in the source frame becomes approximately 0.189 meters.
Understanding how frequency and wavelength are interrelated helps in comprehending how the Doppler Effect influences the properties of sound as perceived by different observers, particularly when they are in motion.

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Most popular questions from this chapter

You are standing at a distance \(D\) from an isotropic point source of sound. You walk \(50.0 \mathrm{~m}\) toward the source and observe that the intensity of the sound has doubled. Calculate the distance \(D\).

A tube \(1.20 \mathrm{~m}\) long is closed at one end. A stretched wire is placed near the open end. The wire is \(0.330 \mathrm{~m}\) long and has a mass of \(9.60 \mathrm{~g} .\) It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

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?

A stationary detector measures the frequency of a sound source that first moves at constant velocity directly toward the detector and then (after passing the detector) directly away from it. The emitted frequency is \(f\). During the approach the detected frequency is \(f_{\text {app }}^{\prime}\) and during the recession it is \(f_{\text {rec. }}^{\prime}\) If \(\left(f_{\text {app }}^{\prime}-f_{\text {rec }}^{\prime}\right) / f=\) \(0.500\), what is the ratio \(v_{s} / v\) of the speed of the source to the speed of sound?

Suppose that the sound level of a conversation is initially at an angry \(70 \mathrm{~dB}\) and then drops to a soothing \(50 \mathrm{~dB}\). Assuming that the frequency of the sound is \(500 \mathrm{~Hz}\), determine the (a) initial and (b) final sound intensities and the (c) initial and (d) final sound wave amplitudes.
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