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JEE ADVANCED 2025
Paper-2 2025
Question
An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length $8 m$ with their pivots well separated along the $X$ axis. They are pulled from the equilibrium position in opposite directions along the $X$ axis by a small angular amplitude $\theta_0=\cos ^{-1}(0.9)$ and released simultaneously. If the natural frequency of the transmitter is 660 Hz and the speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ) is $\_\_\_\_$ .
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Solution
$\begin{aligned} & L=8 \mathrm{~m} \quad \theta_0=\cos ^{-1}(0.9) f_0=660 \mathrm{~Hz} \\ & V_s=330 \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2\end{aligned}$

$\begin{aligned} & v=\sqrt{2 g l(1-\cos \theta)} \\ & =\sqrt{2 \times 10 \times 8 \times 0.1}=4 \mathrm{~m} / \mathrm{s} \\ & f_{\max }=f_0\left(\frac{v_s+v}{v_s-v}\right) \\ & f_{\min }=f_0\left(\frac{v_s-v}{v_s+v}\right) \\ & f_{\max }-f_{\min }=f_0\left(\frac{v_s+v}{v_s-v}-\frac{v_s-v}{v_s+v}\right) \\ & =f_0\left(\frac{4 \times v_s \times v}{\left(v_s^2-v^2\right)}\right)=\frac{660 \times 4 \times 330 \times 4}{\left(330^2-4^2\right)} \approx 32 \mathrm{~Hz} \\ & \Delta f=32 \mathrm{~Hz}\end{aligned}$
Question Tags
JEE Advance
Physics
Easy
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