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Question #42855

Difficulty: 8/10 | 🟢 Free

The Question

A rigid platform is moving along the $+x$ direction through still air. The speed of sound in still air is $c$ . A sound detector $D$ is mounted at the front of the platform. Two point sound sources, $S_1$ and $S_2$ , are mounted on the platform at distances $L_1$ and $L_2$ respectively behind the detector ( $L_1 > L_2$ ). Both sources emit sound isotropically at a constant frequency $f_0$ in their instantaneous rest frame. The platform starts from rest at $t=0$ and moves with a constant acceleration $a$ . At a later time $t$ , the detector has a velocity $v$ ( $v < c$ ). Assume the sound received by the detector at this instant was emitted after $t=0$ , and that $\frac{aL_1}{(c-v)^2} \ll 1$ . Which of the following expressions best represents the beat frequency recorded by the detector at this instant?

Options

A.
$0$
B.
$f_0 \frac{a(L_1 - L_2)}{c^2}$
C.
$f_0 \frac{a(L_1 - L_2)}{c(c-v)}$
D.
$f_0 \frac{a(L_1 - L_2)}{(c-v)^2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Superposition of Non-identical Waves|Wave Optics - Class 12|Accelerated Frame Hydrostatics

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze wave propagation and the Doppler effect in an accelerated reference frame.

A common trap is assuming that because the source and detector are rigidly connected, their relative velocity is zero, leading to a static phase difference $\Delta \phi = k\Delta x$ and a beat frequency of zero (Option A).

Another trap is calculating the wave transit time using the static distance $L/c$ , which ignores the platform's velocity during the wave's flight and incorrectly yields Option B.

The correct physical framework requires recognizing that the wave propagates in the stationary ground frame (air), meaning the transit time is stretched to $L/(c-v)$ .

Consequently, the source's velocity at the retarded time of emission is significantly lower than the detector's velocity at reception, creating a distance-dependent Doppler shift that manifests as a non-zero, chirping beat frequency.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Waves - Class 11 Part 2
Concept	Superposition of Non-identical Waves
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free