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Physics-

General

Easy

Question

Two loudspeakers L₁ and L₂ 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 330 ms^–¹ then the frequency at which the first maximum is observed is

165 Hz
330 Hz
496 Hz
660 Hz

The correct answer is: 330 Hz

Path difference between the wave reaching at D
$capital delta x equals L subscript 2 end subscript P minus L subscript 1 end subscript P equals square root of 4 0 to the power of 2 end exponent plus 9 to the power of 2 end exponent end root minus 40 equals 41 minus 40$ = 1m
For maximum $capital delta x equals left parenthesis 2 n right parenthesis fraction numerator lambda over denominator 2 end fraction$
For first maximum (n = 1) Þ $1 equals 2 left parenthesis 1 right parenthesis fraction numerator lambda over denominator 2 end fraction$ Þ $lambda equals 1 m$
Þ $n equals fraction numerator v over denominator lambda end fraction equals 330 H z$ .

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