JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Waves & Sound

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

)

Answer: 600

Solution

A

the recorder is equidistant from the two speakers, so the path difference is zero; on reaching

B

it has passed through ten maxima, so the path difference there is

10 λ

. From the geometry,

L_{1} B = 4 0^{2} + 2 0^{2} = 205 = 44.6 m

and

L_{2} B = 4 0^{2} + 3 0^{2} = 50 m

, giving a path difference

50 - 44.6 = 5.4 m

. Thus

10 λ = 5.4 \Rightarrow λ = 0.54 m

, and the frequency is

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

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer

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JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Waves & Sound

Solution

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