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 \(2000.0 \mathrm{~Hz}\). 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?

Step by step solution

01

Identify the given values

We are given: - Emitted frequency, \(f = 2000.0 \mathrm{~Hz}\) - Speed of sound in water, \(v = 1500.0 \mathrm{~m/s}\) - Speed of submarine X (source), \(v_s = 10.0 \mathrm{~m/s}\) - Speed of submarine Y (observer), \(v_o = 15.0 \mathrm{~m/s}\)

02

Determine the formula for the Doppler effect

As both submarines are approaching each other, we'll use the "+" sign in the Doppler effect formula: $$f' = f\frac{v + v_o}{v - v_s}$$

03

Calculate the detected frequency

Plug the given values into the Doppler effect formula: $$f' = 2000.0 \times \frac{1500.0 + 15.0}{1500.0 - 10.0}$$ Solve for \(f'\): $$f' \approx 2027.80 \mathrm{~Hz}$$ So, the frequency detected by submarine Y is approximately \(2027.80 \mathrm{~Hz}\). b) What is the frequency detected by submarine X for the sonar wave reflected off submarine Y?

04

Apply the Doppler effect twice

Since the wave is reflected, we'll apply the Doppler effect formula twice. First, find the frequency emitted by submarine Y after reflecting the wave: $$f_Y = f'\frac{v - v_s}{v + v_o}$$ Then, find the frequency detected by submarine X: $$f_X = f_Y\frac{v - v_o}{v + v_s}$$

05

Calculate the detected frequency

Calculate the frequency emitted by submarine Y: $$f_Y = 2027.80 \times \frac{1500.0 - 10.0}{1500.0 + 15.0} \approx 2000.0 \mathrm{~Hz}$$ Calculate the frequency detected by submarine X: $$f_X = 2000.0 \times \frac{1500.0 - 15.0}{1500.0 + 10.0} \approx 1962.33 \mathrm{~Hz}$$ So, the frequency detected by submarine X is approximately \(1962.33 \mathrm{~Hz}\). 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?

06

Apply the Doppler effect formula for moving away

Now that the submarines are moving away from each other, we'll use the '-' sign in the Doppler effect formula: $$f' = f\frac{v - v_o}{v + v_s}$$

07

Calculate the detected frequency and Doppler shift

Calculate the frequency detected by submarine Y: $$f' = 2000.0 \times \frac{1500.0 - 15.0}{1500.0 + 10.0} \approx 1972.33 \mathrm{~Hz}$$ Calculate the Doppler shift: $$\Delta f = f' - f \approx 1972.33 - 2000 \approx -27.67 \mathrm{~Hz}$$ So, the frequency detected by submarine Y when moving away is approximately \(1972.33 \mathrm{~Hz}\), and the Doppler shift is approximately \(-27.67 \mathrm{~Hz}\).

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