JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

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 equidistance to both loud speakers, is moved by 25 m along the line $A B$ 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 m/s and $5 = 2.23$ )

Your answer

Solution

Set up coordinates with $L_{1}$ at $(0, 5)$ , $L_{2}$ at $(0, - 5)$ , $A$ at $(40, 0)$ (on the perpendicular bisector), and $B$ at $(40, 25)$ .

Distances from B to each speaker:

$L_{1} B = 4 0^{2} + (25 - 5)^{2} = 1600 + 400 = 2000 = 205 \approx 44.6$ m.

$L_{2} B = 4 0^{2} + (25 + 5)^{2} = 1600 + 900 = 2500 = 50$ m.

Path difference at B: $Δ x = L_{2} B - L_{1} B = 50 - 44.6 = 5.4$ m.

10 cycles of minima/maxima from A (where $Δ x = 0$ ) to B means the path difference at B is $10 λ$ : $5.4 = 10 λ$ , so $λ = 0.54$ m.

Frequency: $f = \frac{v}{λ} = \frac{324}{0.54} = 600$ Hz.

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