First slide

Superposition of waves

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

Moderate

A

165 Hz
B

330 Hz
C

496 Hz
D

660 Hz

Solution

Path difference between the wave reaching at D
$Δx = L_{2} P - L_{1} P = \sqrt{40^{2} + 9^{2}} - 40 = 41 - 40 = 1 m$
For maximum $Δx = (2 n) \frac{λ}{2}$
For first maximum $(n = 1) \Rightarrow 1 = 2 (1) \frac{λ}{2} \Rightarrow λ = 1 m$
$\Rightarrow n = \frac{v}{λ} = 330 Hz$

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