A source (S) of sound has frequency 240 Hz. When the observer (O) and the source move towards each other at a speed with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be 288 Hz. [2024]

Question

A source (S) of sound has frequency 240 Hz. When the observer (O) and the source move towards each other at a speed $v$ with respect to the ground (as shown in Case 1 in the figure), the observer measures the frequency of the sound to be 288 Hz. However, when the observer and the source move away from each other at the same speed $v$ with respect to the ground (as shown in Case 2 in the figure), the observer measures the frequency of sound to be $n$ Hz. The value of $n$ is _____. [2024]

Smart Achievers · Accepted Answer

(200)

$Using Doppler's effect, frequency received by observer$

$n$

$Case-1: f_{1} = (\frac{C + V}{C - V}) f_{s}$

$\Rightarrow 288 = (\frac{C + V}{C - V}) 240 .....(i)$

$Case-2: f_{2} = (\frac{C - V}{C + V}) f_{s}$

$or, n = (\frac{C - V}{C + V}) 240 .....(ii)$

Multiplying eq. (i) & (ii)

$(288) (n) = (240) (240)$

$∴$ $n = \frac{240 \times 240}{288} = 200$

Solution

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Solution

Ans. (200) Using Doppler's effect, frequency received by observer f0=(C±V0C±Vs)fs, C=speed of sound Case-1: f1=(C+VC-V)fs ⇒ 288=(C+VC-V)240 .....(i) Case-2: f2=(C-VC+V)fs or, n=(C-VC+V)240 .....(ii) Multiplying eq. (i) & (ii) (288)(n)=(240)(240) ∴ n=240×240288=200

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