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Chapter 10: Problem 1462

A rocket is moving at a speed of \(130 \mathrm{~m} / \mathrm{s}\) towards a stationary target. While moving, it emits a wave of frequency $800 \mathrm{~Hz}$. Calculate the frequency of the sound as detected by the target. (Speed of wave \(=330 \mathrm{~m} / \mathrm{s}\) ) (A) \(1320 \mathrm{~Hz}\) (B) \(2540 \mathrm{~Hz}\) (C) \(1270 \mathrm{~Hz}\) (D) \(660 \mathrm{~Hz}\)

Short Answer

Expert verified
The frequency of the sound as detected by the target is (A) \(1320 \mathrm{~Hz}\).

Step by step solution

01

Understand the Doppler effect formula for sound waves

The Doppler effect formula for sound waves is given by: \[f' = f \frac{v + v_0}{v + v_s}\] Here, \(f'\) = Apparent frequency detected by the target \(f\) = Source frequency (800 Hz) \(v\) = Speed of the sound wave (330 m/s) \(v_0\) = Speed of the target (0 m/s, as it is stationary) \(v_s\) = Speed of the source (the rocket), (positive when the source moves away from the observer and negative when it moves towards)
02

Plug the given values into the formula

We know the source frequency, speed of the sound wave, and speed of the rocket, so we can plug these values into the formula. Since the rocket is moving towards the target, the speed of the source will be negative. Hence, \(v_s = -130 \mathrm{~m / s}\).
03

Calculate the apparent frequency

Now, we can plug all these values into the Doppler effect formula and find the apparent frequency detected by the target. \[f' = 800 Hz \frac{330 m/s + 0}{330 m/s - 130 m/s}\]
04

Simplify and find the answer

Simplifying the equation, we get \[f' = 800 Hz \frac{330 m/s}{200 m/s}\] \[f' = 800 Hz \cdot \frac{33}{20}\] \[f' = 1320 Hz\] So, the frequency of the sound as detected by the target will be \(1320 Hz\). Therefore, the correct answer is (A) \(1320 \mathrm{~Hz}\).

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Most popular questions from this chapter

For a particle executing \(\mathrm{S} . \mathrm{H} \mathrm{M} .\), when the potential energy of the oscillator becomes \(1 / 8\) the maximum potential energy, the displacement of the oscillator in terms of amplitude A will be........... (A) \((\mathrm{A} / \sqrt{2})\) (B) \(\\{\mathrm{A} /(2 \sqrt{2})\\}\) (C) \((\mathrm{A} / 2)\) (D) \(\\{\mathrm{A} /(3 \sqrt{2})\\}\)

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