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VITEEE 2009 Paper

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Question : 18 of 120

Marks: +1, -0

Two sources A and B are sending notes of frequency 680 Hz. A listener moves from A and B with a constant velocity u. If the speed of sound in air is

340\,\mathrm{ms}^{-1}

, what must be the value of u so that he hears 10 beats pers second ?

$2.0\; \mathrm{ms}^{-1}$
$2.5\; \mathrm{ms}^{-1}$
$3.0\; \mathrm{ms}^{-1}$
$3.5\; \mathrm{ms}^{-1}$

Solution:

Let listener go from A → B with velocity (µ).

And the apparent frequency of sound from source A by listener using Doppler's effect,

n' = n \frac{v - v_o}{v + v_s}

⇒

n' = 680 \frac{340 - u}{340 + 0}

The apparent frequency of sound from source B by listener

n^{n} = n \frac{v + v_o}{v - v_s}

= 680 \frac{340 + u}{340 - 0}

Listener hear 10 beats per second.

Hence, n'' - n' = 10

\Rightarrow 680 \frac{340 + u}{340} - 680 \frac{340 - u}{340} = 10

⇒ 2(340 + u - 340 + u ) = 10 ⇒ u = 2.5

\mathrm{ms}^{-1}

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