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Chapter 12: Problem 49

A bat emits chirping sounds of frequency \(82.0 \mathrm{kHz}\) while hunting for moths to eat. If the bat is flying toward the moth at a speed of $4.40 \mathrm{m} / \mathrm{s}\( and the moth is flying away from the bat at \)1.20 \mathrm{m} / \mathrm{s},$ what is the frequency of the sound wave reflected from the moth as observed by the bat? Assume it is a cool night with a temperature of \(10.0^{\circ} \mathrm{C} .\) [Hint: There are two Doppler shifts. Think of the moth as a receiver, which then becomes a source as it "retransmits" the reflected wave. \(]\)

Short Answer

Expert verified
Answer: The frequency of the reflected sound wave is approximately 87,043.52 Hz.

Step by step solution

01

Calculate the speed of sound in air

The speed of sound in air depends on the temperature, and we are given that the temperature is \(10.0^{\circ} \mathrm{C}\). We can find the speed of sound using the formula: \(v_{sound} = 331.4 + 0.6T\) where \(T\) is the temperature in Celsius. Plugging the given temperature, we get: \(v_{sound} = 331.4 + 0.6(10.0) = 337.4 \, \mathrm{m/s}\).
02

Calculate the Doppler shift for the first interaction (bat to moth)

Now consider the first Doppler shift when the bat's chirping reaches the moth. Let \(f_{1}\) be the frequency observed by the moth. Since the bat is moving towards the moth, and the moth is moving away from the bat, we use the following formula for the Doppler effect: \(f_{1} = (\frac{v_{sound} + v_{observer}}{v_{sound} - v_{source}})f_{source}\). Here, \(v_{source}= 4.40 \, \mathrm{m/s}\) and \(v_{observer}= 1.20\, \mathrm{m/s}\). So, \(f_{1} = (\frac{337.4+ 1.20}{337.4 - 4.40})82,000 \approx 84,509.12 \, \mathrm{Hz}\).
03

Calculate the Doppler shift for the second interaction (moth to bat)

Now consider the second Doppler shift when the moth reflects the sound back to the bat. Since the bat is now the observer and moth is a source, we must switch the source and observer velocities. The formula for the second Doppler effect will be: \(f_{reflected} = (\frac{v_{sound} + v_{observer}}{v_{sound} - v_{source}})f_{1}\). Here, \(v_{source}= 1.20 \, \mathrm{m/s}\) and \(v_{observer}= 4.40\, \mathrm{m/s}\), and \(f_{1}\) is the frequency calculated in step 2. So, \(f_{reflected} = (\frac{337.4 + 4.40}{337.4 - 1.20})84,509.12 \approx 87,043.52\, \mathrm{Hz}\).
04

Report the frequency of the reflected sound wave

The frequency of the sound wave reflected from the moth, as observed by the bat, is approximately \(87,043.52 \, \mathrm{Hz}\).

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Most popular questions from this chapter

In this problem, you will estimate the smallest kinetic energy of vibration that the human ear can detect. Suppose that a harmonic sound wave at the threshold of hearing $\left(I=1.0 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\right)\( is incident on the eardrum. The speed of sound is \)340 \mathrm{m} / \mathrm{s}\( and the density of air is \)1.3 \mathrm{kg} / \mathrm{m}^{3} .$ (a) What is the maximum speed of an element of air in the sound wave? [Hint: See Eq. \((10-21) .]\) (b) Assume the eardrum vibrates with displacement \(s_{0}\) at angular frequency \(\omega ;\) its maximum speed is then equal to the maximum speed of an air element. The mass of the eardrum is approximately \(0.1 \mathrm{g} .\) What is the average kinetic energy of the eardrum? (c) The average kinetic energy of the eardrum due to collisions with air molecules in the absence of a sound wave is about \(10^{-20} \mathrm{J}\) Compare your answer with (b) and discuss.
An organ pipe that is open at both ends has a fundamental frequency of $382 \mathrm{Hz}\( at \)0.0^{\circ} \mathrm{C}$. What is the fundamental frequency for this pipe at \(20.0^{\circ} \mathrm{C} ?\)
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 .)
Some bats determine their distance to an object by detecting the difference in intensity between cchoes.(a) If intensity falls off at a rate that is inversely proportional to the distance squared, show that the echo intensity is inversely proportional to the fourth power of distance. (b) The bat was originally \(0.60 \mathrm{m}\) from one object and \(1.10 \mathrm{m}\) from another. After flying closer, it is now \(0.50 \mathrm{m}\) from the first and at \(1.00 \mathrm{m}\) from the second object. What is the percentage increase in the intensity of the ccho from each object?
(a) What is the pressure amplitude of a sound wave with an intensity level of \(120.0 \mathrm{dB}\) in air? (b) What force does this exert on an eardrum of area \(0.550 \times 10^{-4} \mathrm{m}^{2} ?\)
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