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 [image:15_image25.png]

Question

Dr. Abubacker · Accepted Answer

(b) 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 $(n = 1) \Rightarrow 1 = 2(1)\frac{\lambda}{2} \Rightarrow \lambda = 1m$
$\Rightarrow n = \frac{v}{\lambda} = 330Hz$

Explanation

Reviewed by Dr. Abubacker

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