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← JEE Advanced 2025 · Paper 2
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JEE Advanced 2025 · Paper 2, Physics question 16

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 XX axis. They are pulled from the equilibrium position in opposite directions along the XX axis by a small angular amplitude θ0=cos⁡−1(0.9)\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 m/s, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity g=10 m/s2g = 10\ \mathrm{m/s^2}) is _____.

Two simple pendulums hanging from a common ceiling, pivots separated along X (X to the right, Y up, g down). The transmitter T on the left pendulum is pulled to the left by angle θ₀ from its vertical; the receiver R on the right pendulum is pulled to the right by θ₀.
Answer and worked solution

Official answer: 26

Maximum speed of each pendulum.
v=2gL(1−cos⁡θ0)=2⋅10⋅8⋅0.1=4 m/s.v=\sqrt{2gL(1-\cos\theta_0)}=\sqrt{2\cdot10\cdot8\cdot0.1}=4\ \text{m/s}.
The two swing as mirror images, so at the bottom they move directly towards or away from each other, each at 4 m/s.

Doppler extremes.

  • Approaching: f=660⋅330+4330−4=676.2f=660\cdot\frac{330+4}{330-4}=676.2 Hz.
  • Receding: f=660⋅326334=644.2f=660\cdot\frac{326}{334}=644.2 Hz.

Maximum variation. ≈32\approx32 Hz.