Two loudspeakers (L₁ and L₂) 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 √5=2.23 )

Solution

Physics sir · 4 steps

Each full max-to-max cycle means the path difference grew by one wavelength, so find the path difference at B.

1. Set up coordinates. Put the midpoint of the speakers at the origin, so L₁ is at (0, 5) and L₂ is at (0, -5). Point A is at (40, 0) and B is 25 m up the line, at (40, 25).
2. Path difference at B. At A both distances are equal, so the path difference is zero. At B: L₂B = √(40² + 30²) = 50 m and L₁B = √(40² + 20²) = 20√5 = 44.6 m.

   Δ x = L₂B - L₁B = 50 - 44.6 = 5.4 m

3. Count wavelengths. One complete cycle of maximum and minimum means the path difference changed by one wavelength. Ten cycles give 10λ = 5.4 m, so λ = 0.54 m.
4. So. f = v/λ = 324/0.54 = 600 Hz, so the answer is 600.

The common slip: 300. Treating each cycle as a half-wavelength change in path difference gives 5λ = 5.4 m and a frequency of 300 Hz, not 600.

The idea to keep. As a detector moves past two coherent sources, every full cycle of loud and soft is one extra wavelength of path difference. Find the change in path difference and divide by the number of cycles.

Answer from the NTA final answer key. Solution by PrepWiser.