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Chapter 16: Problem 71

You are riding your bicycle directly away from a stationary source of sound and hear a frequency that is \(1.0 \%\) lower than the emitted frequency. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is your speed?

Short Answer

Expert verified
Your speed is approximately 3.47 m/s.

Step by step solution

01

Understand the Doppler Effect Formula

The Doppler Effect for sound describes how the frequency of sound changes for an observer moving relative to the sound source. The formula for the observed frequency \(f'\) when moving away is given by: \[ f' = \frac{f}{1 + \frac{v_o}{v}} \] where \(f'\) is the observed frequency, \(f\) is the emitted frequency, \(v_o\) is the speed of the observer, and \(v\) is the speed of sound.
02

Interpret the Given Information

You are given that the frequency you hear is 1.0% lower than the emitted frequency. This means \(f' = 0.99f\). You also know the speed of sound \(v = 343\, \rm{m/s}\).
03

Substitute and Calculate

Substitute \(f' = 0.99f\) and \(v = 343 \mathrm{~m/s}\) into the Doppler Effect formula: \[ 0.99f = \frac{f}{1 + \frac{v_o}{343}} \] Simplifying, we get: \[ 0.99 = \frac{1}{1 + \frac{v_o}{343}} \].
04

Solve for the Observer's Speed

Rearrange the equation to solve for \(v_o\): \[ 1 + \frac{v_o}{343} = \frac{1}{0.99} \]. Then, \[ \frac{v_o}{343} = \frac{1}{0.99} - 1 \]. Finally, \[ v_o = 343 \left(\frac{1}{0.99} - 1\right) \]. Calculate \(v_o\) to find the observer's speed.
05

Perform the Calculation

Calculate \( \frac{1}{0.99} - 1 \approx 0.010101 \). So, \[ v_o = 343 \times 0.010101 \approx 3.47 \mathrm{~m/s} \].

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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 the number of sound wave cycles that pass a specific point per second. It's measured in Hertz (Hz). The frequency determines the pitch of the sound that we hear鈥攈igher frequencies correspond to higher pitches. In our exercise, the frequency that the observer hears is not exactly the original frequency being emitted by the sound source due to motion. This change in frequency that the observer receives compared to what was initially emitted is central to understanding the Doppler Effect. When moving away from the source, the frequency decreases, which is why the observer hears a lower pitch compared to the actual sound that was emitted.
Observer Moving Relative to Sound
When an observer moves relative to the source of sound, fascinating changes happen to the frequency they perceive. If the observer is moving towards the source, the frequency increases, leading to a higher pitch. Conversely, if moving away, as in our problem, the frequency decreases, resulting in a lower pitch.

This shift in frequency happens because, as the observer moves away, the sound waves are stretched out, leading to fewer waves being encountered per second. It's important to note that this change in perceived frequency is a result of the relative motion between the source and observer, and not any change in the speed of sound or the original sound frequency itself.
Speed of Sound
The speed of sound is a crucial constant in problems related to the Doppler Effect. It represents how quickly sound waves travel through a medium, typically air in our context. It is typically around 343 m/s in air at room temperature. Variations occur due to factors like temperature, humidity, and the medium through which the sound is traveling.

Understanding that the speed of sound is constant in a given medium allows us to use it as a reliable reference point when calculating changes in perceived frequency based on the observer's speed. Unlike the frequency, which changes based on observed conditions and movement, the speed of sound remains unaffected by the observer's motion.
Doppler Effect Formula
The Doppler Effect formula is a mathematical representation that helps explain the shift in frequency observed due to the motion of an observer relative to the sound source. For an observer moving away from a stationary sound source, the formula is: \[ f' = \frac{f}{1 + \frac{v_o}{v}} \]

Where:
  • \( f' \) is the observed frequency.
  • \( f \) is the emitted frequency.
  • \( v_o \) is the speed of the observer.
  • \( v \) is the speed of sound.
In our example, the observer hears a frequency 1% lower than the emitted frequency due to moving away from the source. This creates a situation where the formula can be rearranged to solve for the observer's speed, allowing us to directly calculate how fast the observer is moving away based solely on the change in perceived frequency.

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Most popular questions from this chapter

The right-most key on a piano produces a sound wave that has a frequency of \(4185.6 \mathrm{~Hz}\). Assuming that the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\), find the corresponding wavelength.

A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor is a stationary 440 -Hz source of sound. The microphone vibrates up and down in simple harmonic motion with a period of \(2.0 \mathrm{~s}\). The difference between the maximum and minimum sound frequencies detected by the microphone is \(2.1 \mathrm{~Hz}\). Ignoring any reflections of sound in the room and using \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound, determine the amplitude of the simple harmonic motion.

The wavelength of a sound wave in air is \(2.74 \mathrm{~m}\) at \(20{ }^{\circ} \mathrm{C}\). What is the wavelength of this sound wave in fresh water at \(20{ }^{\circ} \mathrm{C} ?\) (Hint: The frequency of the sound is the same in both media.)

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{~m}\) and a mass per unit length of \(0.020 \mathrm{~kg} / \mathrm{m}\). However, the tension in wire \(\mathrm{A}\) is \(6.00 \times 10^{2} \mathrm{~N}\), and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{~N}\). Transverse wave pulses are generated simultaneously, one at the left end of wire \(A\) and one at the right end of wire B. The pulses travel toward each other. How much time does it take until the pulses pass each other?

A bat emits a sound whose frequency is \(91 \mathrm{kHz}\). The speed of sound in air at \(20.0{ }^{\circ} \mathrm{C}\) is \(343 \mathrm{~m} / \mathrm{s}\). However, the air temperature is \(35{ }^{\circ} \mathrm{C}\), so the speed of sound is not \(343 \mathrm{~m} / \mathrm{s} .\) Find the wavelength of the sound.
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