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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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