An observer moves towards a stationary source of sound with a velocity of one-fifth of the velocity of sound…

An observer moves towards a stationary source of sound with a velocity of one-fifth of the velocity of sound. The percentage increase in the apparent frequency is
  1. $5 \%$
  2. $10 \%$
  3. $20 \%$
  4. $25 \%$

Solution

When observer moves towards the stationary source then apparent frequency, $\mathrm{n}^{\prime}=\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}}\right) \mathrm{n}$
Given, $\mathrm{v}_{\mathrm{o}}=\mathrm{v} / 5$. $\therefore \quad \mathrm{n}^{\prime}=\left(\frac{\mathrm{v}+\frac{\mathrm{v}}{5}}{\mathrm{v}}\right) \mathrm{n}=\frac{6}{5} \mathrm{n}$
Increase in apparent frequency $=\mathrm{n}^{\prime}-\mathrm{n}=\frac{6}{5} \mathrm{n}-\mathrm{n}=\frac{1}{5} \mathrm{n} \Rightarrow 20 \%$ of n

Asked in: MHT CET 2024 (03 May Shift 2)

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