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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Waves and Sound

BITSAT 2016 · Physics · Waves and Sound

A source of sound S emitting waves of frequency 100 Hz and an...

A source of sound S emitting waves of frequency 100Hz100 \mathrm{ Hz} and an observor O\mathrm{O} are located at some distance from each other. The source is moving with a speed of 19.4ms−119.4 \mathrm{ ms}^{-1} at an angle of 60∘60^{{}^{\circ}} with the source observer line as shown in the figure. The observor is at rest. The apparent frequency observed by the observer is (velocity of sound in air 330ms−1330 \mathrm{ ms}^{-1} ) A source of sound S emitting waves of frequency 100 Hz and an observor O are located at some distance from each other.

Options

  1. A103Hz103 \mathrm{ Hz}
  2. B106Hz106 \mathrm{ Hz}
  3. C97Hz97 \mathrm{ Hz}
  4. D100Hz100 \mathrm{ Hz}

Correct answer

A. 103Hz103 \mathrm{ Hz}

Step-by-step solution

Here, original frequency of sound, f0=100Hz\mathrm{f}_{0}=100 \mathrm{ Hz} Speed of source Vs=19.4cos⁡60∘=9.7V_{s}=19.4 \cos 60^{{}^{\circ}}=9.7 A source of sound S emitting waves of frequency 100 Hz and an observor O are located at some distance from each other.
From Doppler's formula \mathrmfl=f0(V−V0V−Vs)\mathrmfl=100(V−0V−(+9.7))\mathrmfl=100VV(1−9.7V)\mathrmf1=100(1+9.7330)=103Hz\mathrmf^{\mathrm{l}}=\mathrm{f}_{0}\left(\frac{\mathrm{V}-\mathrm{V}_{0}}{\mathrm{ V}-\mathrm{V}_{\mathrm{s}}}\right)\mathrmf^{\mathrm{l}}=100\left(\frac{\mathrm{V}-0}{\mathrm{ V}-(+9.7)}\right)\mathrmf^{\mathrm{l}}=100 \frac{\mathrm{V}}{\mathrm{V}\left(1-\frac{9.7}{\mathrm{ V}}\right)}\mathrmf^{1}=100\left(1+\frac{9.7}{330}\right)=103 \mathrm{ Hz} Apparent frequency \mathrmf1=103Hz\mathrmf^{1}=103 \mathrm{ Hz}
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