Two loudspeakers (L₁ and L₂) are placed with a separation of 10 m, as… | JEE Main 2026 (22 Jan, Shift 1)

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Reflection, Superposition & Interference

Two loudspeakers $(L_1$ and $L_2)$ are placed with a separation of 10 m, as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point $A$ , at equidistant to both loudspeakers, is moved by 25 m along the line $AB$ while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is ______ Hz (Speed of sound in air is $324\text{ m/s}$ and $\sqrt{5} = 2.23$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 600

At $A$ the path difference is $0$ . At $B$ (25 m up): $L_1B = \sqrt{40^2 + 20^2} = 20\sqrt{5} = 44.6\text{ m}$ , $L_2B = \sqrt{40^2 + 30^2} = 50\text{ m}$ .
Path difference at $B$ $= 50 - 44.6 = 5.4\text{ m}$ .
10 cycles of maxima and minima means the path difference changed by $10\lambda$ : $\lambda = 0.54\text{ m}$ .
$f = \frac{v}{\lambda} = \frac{324}{0.54} = 600\text{ Hz}$ .

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