JEE Advanced 2025 — Paper 2 — PHY-III solution

JEE Advanced2025Paper 2PHY-III

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/s $^{2}$ ) is ___

Your answer

Solution

The small-amplitude approximation gives $cos θ_{0} = 1 - θ_{0}^{2} /2 = 0.9$ , so $θ_{0}^{2} = 0.2$ and $θ_{0} = 1/ 5$ rad.

Both pendulums have angular frequency $ω = g / ℓ = 10/8$ rad/s, and each bob's maximum speed is

$v^{'} = θ_{0} g ℓ = \frac{1}{5} 10 \times 8 = \frac{80}{5} = 4$ m/s.

The maximum (minimum) observed frequency occurs when source and observer approach (recede) along the line joining them, each with speed $v^{'}$ :

$f_{m a x} = f \cdot \frac{v + v ^{'}}{v - v ^{'}}, f_{m i n} = f \cdot \frac{v - v ^{'}}{v + v ^{'}}$

The maximum variation is

$Δ f = f_{m a x} - f_{m i n} = f \cdot \frac{( v + v ^{'} ) ^{2} - ( v - v ^{'} ) ^{2}}{v ^{2} - v ^{'2}} = \frac{4 v v ^{'}}{v ^{2} - v ^{'2}} \cdot f$

Plugging in $v = 330$ m/s, $v^{'} = 4$ m/s, $f = 660$ Hz:

$Δ f = \frac{4 \times 330 \times 4}{33 0 ^{2} - 16} \times 660 \approx \frac{16 \times 330 \times 660}{33 0 ^{2}} \approx \frac{16 \times 660}{330} = 32$ Hz.

$∴ Δ f \approx 32.00$ Hz.

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