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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Sound Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Sound Waves

A source of sound moves along the X-axis at a speed of 22 m s^ -1 ,...

A source of sound moves along the X-axis at a speed of 22 m s−122 \text{ m s}^{-1}, continuously emitting a sound of frequency 2.0 kHz2.0 \text{ kHz} that travels in air at a speed of 330 m s−1330 \text{ m s}^{-1}. A listener QQ stands on the Y-axis at a distance of 330 m330 \text{ m} from the origin. At t=0t = 0, the source crosses the origin PP. What will be the frequency heard by the listener at the instant the sound emitted from PP reaches them?

Options

  1. A2.0 kHz2.0 \text{ kHz}
  2. B1.88 kHz1.88 \text{ kHz}
  3. C2.13 kHz2.13 \text{ kHz}
  4. D2.14 kHz2.14 \text{ kHz}

Correct answer

A. 2.0 kHz2.0 \text{ kHz}

Step-by-step solution

The sound received by the listener was emitted when the source was at the origin PP.

At this instant, the velocity of the source is directed along the X-axis, while the line joining the source to the listener lies along the Y-axis.

The angle θ\theta between the velocity vector of the source and the line of sight from the source to the listener is 90∘90^\circ.

The apparent frequency f′f' heard by the listener is given by the Doppler effect formula:

f' = f \left(\dfrac{v}{v - v_s \cos \theta} \right)

where vv is the speed of sound and vsv_s is the speed of the source.

Substituting θ=90∘\theta = 90^\circ, we get cos⁡90∘=0\cos 90^\circ = 0:

f' = f \left(\dfrac{v}{v - 0} \right) =ff

Given the original frequency f=2.0 kHzf = 2.0 \text{ kHz}, the frequency heard by the listener is:

f′=2.0 kHzf' = 2.0 \text{ kHz}
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