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Chapter 12: Problem 38

A source of sound waves of frequency \(1.0 \mathrm{kHz}\) is traveling through the air at 0.50 times the speed of sound. (a) Find the frequency of the sound received by a stationary observer if the source moves toward her. (b) Repeat if the source moves away from her instead.

Short Answer

Expert verified
Answer: When the source of sound is moving towards the observer, the frequency received is \(\frac{2000}{3} \ \mathrm{Hz}\). When the source of sound is moving away from the observer, the frequency received is \(2000 \ \mathrm{Hz}\).

Step by step solution

01

Case (a): Source moving toward the observer

Since the source is moving towards the observer, the velocity of the source \(v_s\) will have the opposite sign to the velocity of sound in the air. Therefore, the source's speed \(v_s = -0.50v\). We can plug this into the Doppler effect formula: \(f_r = f_s \frac{v + v_0}{v - (-0.50v)}\), with \(f_s = 1000 \ \mathrm{Hz}\), \(v_0 = 0\), and \(v_s = -0.50v\) \(f_r = 1000 \frac{v}{v + 0.50v} = 1000 \frac{v}{1.50v}\) Now we can simplify and solve for \(f_r\): \(f_r = 1000 \cdot \frac{1}{1.5} = 1000 \cdot \frac{2}{3} = \frac{2000}{3} \ \mathrm{Hz}\) So the frequency received by the observer when the source is moving towards her is \(\frac{2000}{3} \ \mathrm{Hz}\).
02

Case (b): Source moving away from the observer

In this case, the source is moving away from the observer, so the velocity of the source \(v_s = 0.50v\). We can plug this into the Doppler effect formula: \(f_r = f_s \frac{v + v_0}{v - 0.50v}\), with \(f_s = 1000 \ \mathrm{Hz}\), \(v_0 = 0\), and \(v_s = 0.50v\) \(f_r = 1000 \frac{v}{v - 0.50v} = 1000 \frac{v}{0.50v}\) Now we can simplify and solve for \(f_r\): \(f_r = 1000 \cdot \frac{1}{0.5} = 1000 \cdot 2 = 2000 \ \mathrm{Hz}\) So the frequency received by the observer when the source is moving away from her is \(2000 \ \mathrm{Hz}\).

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Most popular questions from this chapter

Bats emit ultrasonic waves with a frequency as high as $1.0 \times 10^{5} \mathrm{Hz} .$ What is the wavelength of such a wave in air of temperature \(15^{\circ} \mathrm{C} ?\)
At what frequency \(f\) does a sound wave in air have a wavelength of $15 \mathrm{cm},$ about half the diameter of the human head? Some methods of localization work well only for frequencies below \(f\), while others work well only above \(f\). (See Conceptual Questions 4 and 5 .)
(a) What should be the length of an organ pipe, closed at one end, if the fundamental frequency is to be \(261.5 \mathrm{Hz} ?\) (b) What is the fundamental frequency of the organ pipe of part (a) if the temperature drops to \(0.0^{\circ} \mathrm{C} ?\)
A source and an observer are each traveling at 0.50 times the speed of sound. The source emits sound waves at \(1.0 \mathrm{kHz}\). Find the observed frequency if (a) the source and observer are moving towand each other, (b) the source and observer are moving away from each other; (c) the source and observer are moving in the same direction.
A sound wave with an intensity level of \(80.0 \mathrm{dB}\) is incident on an eardrum of area \(0.600 \times 10^{-4} \mathrm{m}^{2} .\) How much energy is absorbed by the eardrum in 3.0 min?
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