Q140 / 160 · TS EAMCET 2023 (14 May Shift 1) · Physics

A source and an observer move away from each other with same velocity of $10 \mathrm{~ms}^{-1}$ with respect to ground. If the observer finds the frequency of sound coming from the source as $1980 \mathrm{~Hz}$, then actual frequency of the source is (speed of sound in air $=340 \mathrm{~ms}^{-1}$ )

  • A
    $1950 \mathrm{~Hz}$
  • Correct answer
    B
    $2100 \mathrm{~Hz}$
  • C
    $2132 \mathrm{~Hz}$
  • D
    $2486 \mathrm{~Hz}$

Solution

By Doppler effect formula $$ f=f_o\left(\frac{v-v_0}{v+v_s}\right) $$ Here $\mathrm{f}=1980 \mathrm{~Hz}$ Speed of sound in air, $v=340 \mathrm{~m} / \mathrm{s}$ $$ \begin{aligned} & v_o=10 \mathrm{~m} / \mathrm{s} \\ & v_s=10 \mathrm{~m} / \mathrm{s} \\ & 1980=\mathrm{f}_o\left(\frac{340-10}{340+10}\right) \\ & 1980=\mathrm{f}_{\mathrm{o}} \times \frac{330}{350} \\ & \mathrm{f}_{\mathrm{o}}=\frac{1980 \times 35}{33}=2100 \mathrm{~Hz} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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