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 m/s, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity \(g=10\) m/s²) is ____

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Waves & Sound

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

θ_{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/s²) is ____

Answer: 32

Solution

From

cos θ_{0} = 0.9

, we get

θ_{0} = \frac{1}{5}

(using small angle approximation).

Maximum velocity of each pendulum:

v^{'} = ℓ ω θ_{0} = ℓ θ_{0} \frac{g}{ℓ} = θ_{0} g ℓ = \frac{1}{5} 10 \times 8 = 4 m/s

Maximum Doppler shift occurs when both move toward each other:

Δ f_{m a x} = \frac{4 v v ^{'}}{v ^{2} - v ^{'2}} f \approx \frac{4 v v ^{'}}{v ^{2}} f = \frac{4 v ^{'} f}{v}

Δ f_{m a x} = \frac{4 \times 4 \times 660}{330} = 32 Hz

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer

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JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Waves & Sound

Solution

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