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Chapter 16: Problem 72

You are flying in an ultra-light aircraft at a speed of \(39 \mathrm{~m} / \mathrm{s}\). An eagle, whose speed is 18 \(\mathrm{m} / \mathrm{s}\), is flying directly toward you. Each of the given speeds is relative to the ground. The eagle emits a shrill cry whose frequency is \(3400 \mathrm{~Hz}\). The speed of sound is \(330 \mathrm{~m} / \mathrm{s}\). What frequency do you hear?

Short Answer

Expert verified
You hear a frequency of approximately 4021 Hz.

Step by step solution

01

Understanding the Doppler Effect

The Doppler effect describes a change in frequency or wavelength of a wave in relation to an observer moving relative to the wave source. Here, both the source (the eagle) and the observer (you in the aircraft) are moving, so we need to apply the Doppler effect formula for both moving source and observer.
02

Define the Variables

Let \( f_s = 3400 \, \text{Hz} \) be the frequency emitted by the eagle, \( v = 330 \, \text{m/s} \) be the speed of sound, \( v_s = 18 \, \text{m/s} \) be the speed of the eagle, and \( v_o = 39 \, \text{m/s} \) be the speed of the observer (you).
03

Use the Doppler Effect Formula

The frequency heard by the observer when both the source and observer are moving towards each other is given by:\[ f' = \frac{v + v_o}{v - v_s} \times f_s \]Substitute \( v = 330 \, \text{m/s} \), \( v_s = 18 \, \text{m/s} \), \( v_o = 39 \, \text{m/s} \), and \( f_s = 3400 \, \text{Hz} \) into the formula.
04

Calculate the Observed Frequency

Plug the values into the formula:\[ f' = \frac{330 + 39}{330 - 18} \times 3400 \]\[ f' = \frac{369}{312} \times 3400 \]\[ f' \approx 1.1827 \times 3400 \]\[ f' \approx 4021 \text{ Hz} \]
05

Conclusion

The frequency that you hear, as the observer in the aircraft, is approximately \( 4021 \, \text{Hz} \).

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Key Concepts

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

Frequency
Frequency is the number of times a wave oscillates or repeats in one second. It is a critical aspect in understanding waves, including sound waves. Frequency is measured in Hertz (Hz), which directly relates to the pitch we hear. High-frequency sounds have a high pitch, while low-frequency sounds have a low pitch.
In this exercise, the eagle's cry has a frequency of 3400 Hz. This frequency tells us the original number of vibrations per second the eagle produces. However, due to the Doppler Effect, the frequency perceived by an observer can change if either the source, the observer, or both are moving.
Wave Source
The wave source is the origin from where the waves propagate. In this scenario, the eagle acts as the wave source, emitting sound waves at a frequency of 3400 Hz. As it moves towards the observer, the distance between each subsequent wavefront decreases.
This movement effectively increases the frequency heard by the observer due to wave compression. Understanding the behavior of the wave source is key to applying the Doppler Effect, which explains the change in frequency due to relative motion.
Sound Speed
Sound speed refers to the speed at which sound waves travel through a medium, such as air. For this problem, the speed of sound is given as 330 m/s, which is typical at room temperature.
Sound speed is critical for calculating the Doppler Effect, as it sets the baseline for how quickly sound waves can reach the observer. In the equation applied, sound speed is used to adjust the relative velocities of the source and observer, which in turn affects the observed frequency. It's important to know that sound speed can vary with changes in conditions like temperature and pressure.
Moving Observer
A moving observer influences the perceived frequency of a sound according to the Doppler Effect. In this case, the observer is flying in an ultra-light aircraft at 39 m/s towards the source of the sound, which is the eagle.
The speed of the observer adds to the speed of sound when determining how quickly sound waves reach the observer. If the observer moves towards the wave source, the observed frequency increases because the observer encounters the wavefronts more frequently. Understanding the role of a moving observer is necessary to compute how the frequency changes, as in this exercise, where the observed frequency is significantly higher than the emitted frequency.

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Most popular questions from this chapter

Review Conceptual Example 3 before starting this problem. A horizontal wire is under a tension of \(315 \mathrm{~N}\) and has a mass per unit length of \(6.50 \times 10^{-3} \mathrm{~kg} / \mathrm{m} .\) A transverse wave with an amplitude of \(2.50 \mathrm{~mm}\) and a frequency of \(585 \mathrm{~Hz}\) is traveling on this wire. As the wave passes, a particle of the wire moves up and down in simple harmonic motion. Obtain (a) the speed of the wave and (b) the maximum speed with which the particle moves up and down.

co Concept Questions A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. (a) Is the tension in the rope greater near the top or near the bottom of the rope, or is the tension the same everywhere along the rope? Why? (b) \(\mathrm{A}\) small-amplitude transverse wave is sent up the rope from the bottom end. Is the speed of the wave greater near the bottom or near the top of the rope, or is the speed of the wave the same everywhere along the rope? Explain. (c) Consider a section of the rope between the bottom end and a point that is a distance \(y\) meters above the bottom. What is the weight of this section? Express your answer in terms of \(m, L, y\), and \(g\) (the acceleration due to gravity). Problem (a) For the rope described in the Concept Questions, derive an expression that gives the speed of the wave on the rope in terms of the distance \(y\) above the bottom end and the acceleration \(g\) due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{~m}\) and \(2.0 \mathrm{~m}\) above the bottom end of the rope Be sure that your answers are consistent with your answer to Concept Question (b).

A transverse wave is traveling on a string. The displacement \(y\) of a particle from its equilibrium position is given by \(y=(0.021 \mathrm{~m}) \sin (25 t-2.0 x) .\) Note that the phase angle \(25 t-2.0 x\) is in radians, \(t\) is in seconds, and \(x\) is in meters. The linear density of the string is \(1.6 \times 10^{-2} \mathrm{~kg} / \mathrm{m} .\) What is the tension in the string?

The speed of a transverse wave on a string is \(450 \mathrm{~m} / \mathrm{s}\), and the wavelength is \(0.18 \mathrm{~m}\). The amplitude of the wave is \(2.0 \mathrm{~mm}\). How much time is required for a particle of the string to move through a total distance of \(1.0 \mathrm{~km}\) ?

The equation \(\beta=(10 \mathrm{~dB}) \log \left(I / I_{0}\right),\) which defines the decibel, can be written in terms of power \(P\) (in watts) rather than intensity \(I\) (in watts/meter \(^{2}\) ). The form \(\beta=(10 \mathrm{~dB}) \log \left(P / P_{0}\right)\) can be used to compare two power levels in terms of decibels. Suppose that stereo amplifier A is rated at \(P=250\) watts per channel, and amplifier B has a rating of \(P_{0}=45\) watts per channel, (a) Expressed in decibels, how much more powerful is A compared to \(\mathrm{B} ?\) (b) Will A sound more than twice as loud as \(\mathrm{B}\) ? Justify your answer.
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