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

Short answer

Answer: When the submarines are approaching each other, the frequency detected by submarine Y is approximately 2027.80 Hz. When they are moving away from each other, the frequency is approximately 1972.33 Hz. The Doppler shift in this scenario is approximately -27.67 Hz.

Step-by-step solution

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), \(v_o = 15.0 \mathrm{~m/s}\)

Determine the formula for the Doppler effect

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

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?

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: $$f_X = f_Y\frac{v - v_o}{v + v_s}$$

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?

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: $$f' = f\frac{v - v_o}{v + v_s}$$

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 \(-27.67 \mathrm{~Hz}\).

Key concepts

Frequency Shift

The frequency shift occurs when there is a relative motion between a source of sound and an observer. This is a core part of the Doppler Effect, which describes how the frequency of a wave changes for an observer moving relative to the source of the wave. When the observer and the source are moving closer together, the observed frequency increases. Conversely, when they move apart, the frequency decreases.

Mathematically, the frequency detected by the observer can be calculated using the Doppler Effect formula:

• Approaching: \( f' = f\frac{v + v_o}{v - v_s} \)
• Receding: \( f' = f\frac{v - v_o}{v + v_s} \)

where \( f \) is the original frequency, \( v \) is the speed of sound in the medium, \( v_o \) is the speed of the observer, and \( v_s \) is the speed of the source.

Sound Waves

Sound waves are vibrations that travel through a medium like air, water, or solids. They are longitudinal waves, meaning that the displacement of the medium is in the same direction as the propagation of the wave. Sound waves require a medium to travel, which is why they cannot propagate in a vacuum.

The speed of sound varies depending on the medium it travels through. In water, it's typically around 1500 meters per second. This plays a significant role in problems involving the Doppler Effect, as the speed of sound affects how much the frequency shifts.

Understanding sound waves helps in solving problems like determining how a sonar wave's frequency changes as two submarines approach or move away from each other.

Sonar Wave

Sonar stands for Sound Navigation and Ranging. It's a technique that uses sound propagation, typically underwater, to navigate, communicate, or detect objects. Sonar systems send out sound waves and listen for echoes, which return from objects, similar to the way radar works but using sound waves rather than electromagnetic ones.

In the context of submarines, a sonar wave is used to "ping" another submarine. The frequency of these waves is crucial for detecting and interpreting the signals accurately. When a sonar wave reflects off an object, the frequency of the wave can be altered due to the Doppler Effect, depending on the relative movements of the submarines and the reflecting objects.

• Active sonar: emits pulses and listens for echoes
• Passive sonar: listens for sounds made by objects

Understanding sonar is essential for problems involving frequency shifts in underwater navigation.