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, equidistant from 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 m/s and 5 = 2.23) [figure]
Answer
600
Solution
Take the mid-point of L1L2 as origin: L1 = (0, 5), L2 = (0, -5), A = (40, 0) and B = (40, 25) (in m).
At A the path difference is zero (central maximum).
At B:
L1B = 402 + 202 = 2000 = 20 5 = 44.6 m
L2B = 402 + 302 = 50 m
x = 50 - 44.6 = 5.4 m
Each cycle of maximum and minimum corresponds to a change of path difference of one wavelength:
10 = 5.4 = 0.54 m
f = v = 3240.54 = 600 Hz
Answer: 600