JEE Advanced 2025 Paper 2 · Q16 · Sound Waves
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 $\theta_0 = \arccos(0.9)$ and released simultaneously. If the natural frequency of the transmitter is $660$ Hz and the speed of sound in air is $330$ m/s, the maximum variation in the frequency (in Hz) as measured by the receiver is __________. (Take $g = 10$ m/s².)
Reveal answer + step-by-step solution
Correct answer:32
Solution
From $\cos(\theta_0) = 0.9$, the maximum speed of each pendulum at the equilibrium position is $v_{\max} = \sqrt{2 g L (1 - \cos\theta_0)} = \sqrt{2 \times 10 \times 8 \times 0.1} = \sqrt{16} = 4$ m/s. When released simultaneously the two bobs swing in opposition; at equilibrium both move with maximum speed in opposite directions, so they approach each other with relative speed $2 v_{\max} = 8$ m/s (maximum approach), or recede at $8$ m/s after they cross. Doppler (both source and observer moving towards each other): $f_{\max} = f (c + v_{\text{obs}})/(c - v_{\text{src}}) = 660(330 + 4)/(330 - 4) = 660 \times 334/326$. $f_{\min} = 660 \times 326/334$. Variation $= f_{\max} - f_{\min} = 660(334^2 - 326^2)/(326 \times 334) = 660(8 \times 660)/108916 \approx 32.0$ Hz.
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