Two loudspeakers `L_1` and `L_2` driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is `330ms^-1` then the frequency at which the first maximum is observed is
A
165 Hz
B
330 Hz
C
495 Hz
D
660 Hz
Text Solution
Verified by Experts
The correct Answer is:
B
Let `Deltax` = path difference `rArr" "Deltax = L_(2)D-L_(1)D rArr Deltax = sqrt(40^(2)+9^(2))-40` `rArr" "Deltax = 41-41 rArr Deltax= 1m` For first maximum , `Deltax = (2n)(lamda)/2 " "(where'n= 1)` `rArr1= 2 (1)(lamda)/2rArrlamda= 1m rArrf=v/lamda=330 Hz`
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