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

Question

Infinity Learn · Accepted Answer

Path difference between the wave reaching at D  Δ$$x=L2P$$−$$L1P=402+92$$−$$40=41$$−$$40=1mFor$$ maximum Δ$$x=(2n)$$λ$$2For$$ first maximum $$n=1$$⇒$$1=2(1)$$λ$$2$$⇒λ$$=1m$$⇒$$n=v$$λ$$=330Hz$$

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