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 $$AB$$ 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 m/s and $$ \sqrt{5}=2.23 $$)

Now we can solve it.

Speakers are separated by

$$L_1L_2=10m$$

so each is 5 m above and below midpoint.

Point A is 40 m from midpoint, and is equidistant from both speakers.

Distance from A to each speaker:

$$L_1A=L_2A=\sqrt{40^2+5^2}$$

At point B, recorder has moved 25 m upward, so coordinates relative to midpoint are

(40,25)

Distance to upper speaker $$L_1$$​:

$$L_1B=\sqrt{40^2+20^2}$$

$$=20\sqrt{5}$$

Distance to lower speaker $$L_2$$​:

$$L_2B=\sqrt{40^2+30^2}$$

$$=50$$

So path difference at BBB is

Using

$$\sqrt{5}=2.235​=2.23$$

Δ=50−44.6=5.4 m

Initially at A,

Δ=0

So change in path difference is

5.4 m

During movement, recorder undergoes 10 cycles of minima and maxima, meaning 10 fringe changes:

$$10\lambda=5.4$$

λ=0.54 m

Frequency

$$f=\frac{v}{λ}$$

$$=\frac{324}{0.54}$$

=600