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Chapter 17: Problem 54

Two loudspeakers are mounted on a merry-go-round whose radius is \(9.01 \mathrm{~m} .\) When stationary, the speakers both play a tone whose frequency is \(100.0 \mathrm{~Hz}\). As the drawing illustrates, they are situated at opposite ends of a diameter. The speed of sound is 343.00 \(\mathrm{m} / \mathrm{s},\) and the merry-go-round revolves once every \(20.0 \mathrm{~s}\). What is the beat frequency that is detected by the listener when the merry-go- round is near the position shown?

Short Answer

Expert verified
The beat frequency detected is approximately 1.65 Hz.

Step by step solution

01

Determine the Angular Velocity of the Merry-Go-Round

The merry-go-round completes one full revolution every 20 seconds. We can calculate the angular velocity \( \omega \) using the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period. Here, \( T = 20 \) s.\[ \omega = \frac{2\pi}{20} = \frac{\pi}{10} \, \mathrm{rad/s} \]
02

Calculate the Tangential Speed of the Speakers

The tangential speed \( v_t \) of each speaker is given by the formula \( v_t = \omega r \), where \( r = 9.01 \, \mathrm{m} \) is the radius. Using \( \omega = \frac{\pi}{10} \, \mathrm{rad/s} \), we find:\[ v_t = \frac{\pi}{10} \times 9.01 = 0.901 \pi \, \mathrm{m/s} \]
03

Determine the Doppler Shift for Each Speaker

For one speaker, the motion is towards the listener, and for the other, it's away. The Doppler effect formula for the frequency heard \( f' \) is \( f' = f \frac{v + v_t}{v} \) for the approaching source and \( f' = f \frac{v - v_t}{v} \) for the receding source, where \( v = 343 \, \mathrm{m/s} \) is the speed of sound and \( f = 100 \, \mathrm{Hz} \) is the emitted frequency.Approaching speaker:\[ f'_\text{approach} = 100 \times \frac{343 + 0.901\pi}{343} \]Receding speaker:\[ f'_\text{recede} = 100 \times \frac{343 - 0.901\pi}{343} \]
04

Approximate and Calculate the Beat Frequency

The beat frequency \( f_\text{beat} \) is the difference between the frequencies detected from the two speakers:\[ f_\text{beat} = |f'_\text{approach} - f'_\text{recede}| = \left| 100 \times \left( 1 + \frac{0.901\pi}{343} \right) - 100 \times \left( 1 - \frac{0.901\pi}{343} \right) \right| \]This simplifies to\[ f_\text{beat} = 100 \times 2 \times \frac{0.901\pi}{343} \approx 1.65 \, \mathrm{Hz} \]

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Key Concepts

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

Understanding Beat Frequency
The concept of beat frequency arises when two sound waves of slightly different frequencies interfere with each other. This interference creates a new sound with a frequency known as the beat frequency. The beat frequency is equal to the absolute difference between the two frequencies of the overlapping waves.
Beat frequency is often easy to detect as it results in a repeating oscillation in amplitude, sometimes described as being similar to a 'wah-wah' effect. It is particularly useful for tuning musical instruments and distinguishing subtle frequency differences.
  • Mathematically, if two frequencies are given by \( f_1 \) and \( f_2 \), then the beat frequency \( f_{\text{beat}} \) is simply \(|f_1 - f_2|\).
  • In the context of the Doppler Effect, when two speakers on a moving platform emit sounds of slightly varying frequencies due to their differing speeds relative to the listener, the listener perceives a beat frequency representing the sound variation.
Exploring Angular Velocity
Angular velocity is a measure of how quickly an object rotates or revolves around a central point. It is an essential parameter for understanding rotational motion.
In our exercise, the merry-go-round's angular velocity determines how fast the speakers are moving along their circular path. It's calculated by the formula:
  • \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of rotation.
  • The unit of angular velocity is radians per second (rad/s).
Understanding angular velocity is crucial when analyzing rotating systems. In this case, it helps us determine the effect of motion on the sound waves produced by the speakers.
Calculating Tangential Speed
Tangential speed refers to the linear speed of any object moving along a circular path. It's particularly relevant to the speakers on this rotating merry-go-round.
The tangential speed can be calculated using the angular velocity derived earlier. The formula is:
  • \( v_t = \omega r \), where \( r \) is the radius of the circle.
  • It represents how fast a point on the edge of the merry-go-round is moving "straight" along the path of the circle.
This value is key when considering the Doppler Effect, as it affects the frequency of the sound waves detected by a listener due to the relative motion between the speakers and the observer.
Analyzing Sound Waves and the Doppler Effect
Sound waves are mechanical waves that propagate through a medium like air. They can be influenced by motion, as described by the Doppler Effect.
When the source of a sound wave moves relative to a listener, the frequency perceived by the listener changes:
  • If the sound source approaches, the frequency increases 鈥� this is known as blue-shift.
  • If it recedes, the frequency decreases 鈥� known as red-shift.
The formula accounting for these changes is:
\[ f' = f \frac{v \pm v_s}{v} \]where:
  • \( f' \) is the perceived frequency,
  • \( f \) is the emitted frequency,
  • \( v \) is the speed of sound, and
  • \( v_s \) is the speed of the source relative to the listener. For each speaker on the merry-go-round, you calculate the perceived frequency based on its direction relative to the listener.
Understanding how sound waves behave when the source or listener is moving is fundamental to resolving scenarios involving the Doppler Effect, such as the one presented in this exercise.

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Most popular questions from this chapter

A string of length \(0.28 \mathrm{~m}\) is fixed at both ends. The string is plucked and a standing wave is set up that is vibrating at its second harmonic. The traveling waves that make up the standing wave have a speed of \(140 \mathrm{~m} / \mathrm{s}\). What is the frequency of vibration?

Standing waves are set up on two strings fixed at each end, as shown in the drawing. The tensions in the strings are the same, and each string has the same mass per unit length. However, one string is longer than the other, (a) Do the waves on the longer string have a larger speed, a smaller speed, or the same speed as those on the shorter string? Justify your answer. (b) Will the longer string vibrate at a higher frequency, a lower frequency, or the same frequency as the shorter string? Provide a reason for your answer. (c) Will the beat frequency produced by the two standing waves increase or decrease if the longer string is increased in length? Why? The two strings in the drawing have the same tension and mass per unit length, but they differ in length by \(0.57 \mathrm{~cm} .\) The waves on the shorter string propagate with a speed of \(41.8 \mathrm{~m} / \mathrm{s},\) and the fundamental frequency of the shorter string is \(225 \mathrm{~Hz}\). Determine the beat frequency produced by the two standing waves.

An organ pipe is open at both ends. It is producing sound at its third harmonic, the frequency of which is \(262 \mathrm{~Hz}\). The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the length of the pipe?

In Concept Simulation \(17.2\) at you can explore the concepts that are important in this problem. A 440.0-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of \(3 \mathrm{~Hz}\) is heard. When the string is tightened, the frequency at which it vibrates increases, and the beat frequency is heard to decrease. What was the original frequency of the guitar string?

A string has a linear density of \(8.5 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\) and is under a tension of \(280 \mathrm{~N}\). The string is \(1.8 \mathrm{~m}\) long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that make up the standing wave.
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