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 $330m{s}^{-1}$, then the frequency at which the first maximum is observed is

• A. 165 Hz
• B. 330 Hz
• C. 496 Hz
• D. 660 Hz

Answer 1

The correct option is B

330 Hz

Path difference between the wave reaching at D

$\Delta x$ = $L_{2}P-L_{1}P$ = $\sqrt{{40}^2+9^2}-40$ = 41 - 40 = 1 m

For maximum $\Delta x$ = $n\lambda$

For first maximum $(n=1)$ $\Rightarrow$ $1$ = $\left(1\right)\lambda$ $\Rightarrow$ $\lambda$ = $1m$

$\Rightarrow$ frequency = $\frac{v}{\lambda }$ = $330Hz$.