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 `330ms^-1` then the frequency at which the first maximum is observed is
A. `165Hz`
B. `330Hz`
C. `496Hz`
D. `660Hz`

Answer 1

Correct Answer - B
Path difference between the wave reaching at D `Delta x=L_2P-L_1P=sqrt(40^2+9^(2))-40=41-40=1m`
For maximum `Delta x=(2n)(lamda)/(2)`
For first maximum `(n=1)implies1=2(1)(lamda)/(2)implieslamda=1m`
`impliesn=(v)/(lamda)=330Hz`