Two sources $A$ and $B$ are sounding notes of frequency $680 \mathrm{~Hz}$. A listener moves from $A$ to $B$ with a constant velocity $u$. If the speed of sound is $340 \mathrm{~ms}^{-1}$, what must be the value of $u$, so that he hears 10 beats per second?

JEE PHYSICS MCQ: WAVES

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Two sources

A

and

B

are sounding notes of frequency

680 \mathrm{~Hz}

. A listener moves from

A

B

with a constant velocity

u

. If the speed of sound is

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

, what must be the value of

u

, so that he hears 10 beats per second?

View answer and explanation

Correct option: B

2.5 \mathrm{~ms}^{-1}

Option B (

2.5 \mathrm{~ms}^{-1}

) is correct because it's the velocity required for the listener to hear 10 beats per second, calculated using the Doppler effect formula.

Practice All Chapter MCQs →

3\,\,kHz

0.5

f\mathrm{ }Hz

V\mathrm{ }m/sec

0.2\,\,V

y = \frac{3}{{{{\left( {2x + 3t} \right)}^2} + 4}},

(– \mathrm{ve}) \,x

y = f(kx +\omega t),

X

0.01\,\,s

5 \mathrm{~m}

245 \mathrm{~g}

1 \mathrm{~kg}

g=980 \mathrm{~cm} \mathrm{~s}^{-2}

T

V