JEE Advance - Physics (2024 - Paper 1 Online - No. 11)
A source (S) of sound has frequency $240 \mathrm{~Hz}$. 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 $288 \mathrm{~Hz}$. 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 $n \mathrm{~Hz}$. The value of $n$ is ___________.
Answer
200
Explanation
Frequency received by observer
$$\mathrm{f_0=\left(\frac{C \pm V_0}{C \pm V_S}\right) f_s, C}$$ is speed of sound
Case-1:
$$\begin{aligned} & \mathrm{f}_1=\left(\frac{\mathrm{C}+\mathrm{V}}{\mathrm{C}-\mathrm{V}}\right) \mathrm{f}_{\mathrm{s}} \\ & 288=\left(\frac{\mathrm{C}+\mathrm{V}}{\mathrm{C}-\mathrm{V}}\right) 240 \end{aligned}$$
Case-2:
$$\begin{aligned} & \mathrm{f}_2=\left(\frac{\mathrm{C}-\mathrm{V}}{\mathrm{C}+\mathrm{V}}\right) \mathrm{f}_{\mathrm{s}} \\ & \mathrm{n}=\left(\frac{\mathrm{C}-\mathrm{V}}{\mathrm{C}+\mathrm{V}}\right) 240 \end{aligned}$$
multiply the two equations, we get.
$$\begin{aligned} & (288)(\mathrm{n})=(240)(240) \\ & \mathrm{N}=200 \end{aligned}$$
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