Question

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 $330{\text{ms}}^{–1}$ then the frequency at which the first maximum is observed is (in $\text{Hz}$)

Answer 1

Path difference between the waves reaching at $D$
$\Delta x=L_{2}D-L_{1}D=\sqrt{{40}^2+9^2}-40=41-40=1\text{m}$

For maxima, $\Delta x=\left(2n\right)\frac{\lambda }{2}$
For first maximum, $n=1\Rightarrow 1=2\left(1\right)\frac{\lambda }{2}\Rightarrow \lambda =1\text{m}$

Hence, frequency of oscillator
$f=\frac{v}{\lambda }=330\text{Hz}$