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 330 ms –1 then the frequency at which the first maximum is observed is
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Path difference between the wave reaching at D
$\Delta x = L_{2,P} - L_{1,P} = \sqrt{40^2 + 9^2} - 40 = 41 - 40$ = 1m
For maximum $\Delta x = (2n) \frac{\lambda}{2}$ For first maximum $(n=1) \Rightarrow 1 = 2(1) \frac{\lambda}{2} \Rightarrow \lambda = 1 \mathrm{m}$ $\Rightarrow n = \frac{v}{\lambda} = 330 \mathrm{Hz}$.
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