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 $330\,ms^{-1}$ then the frequency at which the first maximum is observed is .... $Hz$
Medium
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(b) Path difference between the wave reaching at $D$
$\Delta x = {L_2}P - {L_1}P = \sqrt {{{40}^2} + {9^2}} - 40 = 41 - 40 = 1m$
For maximum $\Delta x = (2n)\frac{\lambda }{2}$
For first maximum $(n = 1)$ ==>$1 = 2(1)\frac{\lambda }{2}$==> $\lambda = 1m$
==> $n = \frac{v}{\lambda } = 330\,Hz$.
$\Delta x = {L_2}P - {L_1}P = \sqrt {{{40}^2} + {9^2}} - 40 = 41 - 40 = 1m$
For maximum $\Delta x = (2n)\frac{\lambda }{2}$
For first maximum $(n = 1)$ ==>$1 = 2(1)\frac{\lambda }{2}$==> $\lambda = 1m$
==> $n = \frac{v}{\lambda } = 330\,Hz$.
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