Question

In a suspense-thriller movie, two submarines, \(X\) and \(Y\), approach each other, traveling at \(10.0 \mathrm{~m} / \mathrm{s}\) and \(15.0 \mathrm{~m} / \mathrm{s}\), respectively. Submarine X "pings" submarine \(Y\) by sending a sonar wave of frequency 2000.0 Hz. Assume that the sound travels at \(1500.0 \mathrm{~m} / \mathrm{s}\) in the water. a) Determine the frequency of the sonar wave detected by submarine Y. b) What is the frequency detected by submarine \(X\) for the sonar wave reflected off submarine Y? c) Suppose the submarines barely miss each other and begin to move away from each other. What frequency does submarine \(Y\) detect from the pings sent by X? How much is the Doppler shift?

Short answer

Answer: The frequency detected by submarine Y after passing X is approximately 1924.20 Hz, while the frequency detected by submarine X is approximately 2159.83 Hz. The Doppler shift is 75.80 Hz.

Step-by-step solution

Part a) Frequency detected by submarine Y

First, let's find the frequency detected by submarine Y (\(f_Y\)) when submarine X pings it. In this case, submarine Y is the observer, and submarine X is the source. Use the Doppler effect formula with \(v_{sound} = 1500 \mathrm{~m} / \mathrm{s}\), \(v_{observer} = -15.0\mathrm{~m} / \mathrm{s}\) (because it's approaching X), and \(v_{source} = -10.0\mathrm{~m} / \mathrm{s}\) : \(f_Y = \frac{2000 \mathrm{Hz}(1500 \mathrm{~m} / \mathrm{s} - 15\mathrm{~m} / \mathrm{s})}{(1500 \mathrm{~m} / \mathrm{s} - 10 \mathrm{~m} / \mathrm{s})} = 2079.58 \mathrm{Hz}\) So the frequency detected by submarine Y is approximately 2079.58 Hz.

Part b) Frequency detected by submarine X

Now, let's find the frequency detected by submarine X (\(f_X\)) when the sonar wave is reflected off submarine Y. In this case, submarine X is the observer, and submarine Y is the source. Use the Doppler effect formula again, with \(v_{observer} = 10.0\mathrm{~m} / \mathrm{s}\) (because it's approaching Y) and \(v_{source} = 15.0\mathrm{~m} / \mathrm{s}\): \(f_X = \frac{2079.58 \mathrm{Hz}(1500 \mathrm{~m} / \mathrm{s} + 10\mathrm{~m} / \mathrm{s})}{(1500 \mathrm{~m} / \mathrm{s} + 15 \mathrm{~m} / \mathrm{s})} = 2159.83 \mathrm{Hz}\) So the frequency detected by submarine X is approximately 2159.83 Hz.

Part c) Frequency detected by submarine Y after passing X

Finally, let's find the frequency detected by submarine Y (\(f_Y'\)) when the submarines barely miss each other and begin to move away from each other. In this case, both the observer and the source are moving away from each other, so we need to change the signs of their velocities: \(v_{observer} = 15.0\mathrm{~m} / \mathrm{s}\) and \(v_{source} = 10.0\mathrm{~m} / \mathrm{s}\): \(f_Y' = \frac{2000 \mathrm{Hz}(1500 \mathrm{~m} / \mathrm{s} + 15\mathrm{~m} / \mathrm{s})}{(1500 \mathrm{~m} / \mathrm{s} + 10 \mathrm{~m} / \mathrm{s})} = 1924.20 \mathrm{Hz}\) So the frequency detected by submarine Y after passing X is approximately 1924.20 Hz.

Part c) The Doppler shift

The Doppler shift is the difference between the source frequency and the observed frequency. In this case, the Doppler shift is given by: Doppler shift = \(2000 \mathrm{Hz} - 1924.20 \mathrm{Hz} = 75.80 \mathrm{Hz}\) So the Doppler shift is 75.80 Hz.