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 :

• A. $165\text{Hz}$
• B. $330\text{Hz}$
• C. $496\text{Hz}$
• D. $660\text{Hz}$

Answer 1

The correct option is B $330\text{Hz}$

Path difference $\Delta x=\sqrt{(9{)}^2+(40{)}^2}-40=1\text{m}$
Hence phase change $=\delta =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }$
For the first maximum $\delta =\frac{2\pi }{\lambda }=2\pi$
hence, $\lambda =1\text{m}$
frequency $=\nu =\frac{v}{\lambda }=330\text{Hz}$