A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms⁻¹. The velocity of sound in air is 300 ms⁻¹. If the person can hear frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear whistle is (a) 15√(2) ms⁻¹ (b) (15)/(√(2)) ms⁻¹ (c) 15 ms⁻¹ (d) 30 ms⁻¹

Answer: (c). The lowest emitted frequency (9500 Hz) must not exceed 10000 Hz after the shift. 10000 = 9500 × (300)/(300 - v) ⇒ 300 - v = 285 So the answer is (c) v = 15 m/s.

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Doppler Effect

A whistle producing sound waves of frequencies $9500\text{ Hz}$ and above is approaching a stationary person with speed $v\text{ ms}^{-1}.$ The velocity of sound in air is $300\text{ ms}^{-1}.$ If the person can hear frequencies upto a maximum of $10{,}000\text{ Hz},$ the maximum value of $v$ upto which he can hear whistle is

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Answer: (c)

The lowest emitted frequency (9500 Hz) must not exceed 10000 Hz after the shift.
$10000 = 9500 \times \dfrac{300}{300 - v} \Rightarrow 300 - v = 285$
So the answer is (c) $v = 15\text{ m/s}$ .

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