As shown in the fig two loudspeakers are located at point A and B. Both are vibrating in phase at a frequency $ν$ and $P_{1}$ and $P_{2}$ are their respective power. Point C lies on a line joining two loudspeakers at a distance of $d_{1}$ from A and $d_{2}$ from B. When both the speakers are switched on, what is the intensity $(i n W / m^{2})$ at point C? Take velocity of sound $300 m s^{- 1},$ frequency $ν$ =100Hz, $d_{1} = 1 m a n d d_{2} = 1.5 m, p_{1} = 8 π W a n d p_{2} = 18 π W .$ Also assume that loudspeakers behave like isotropic point sources: (emit sound uniformly in all directions)

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$6 W / m^{2}$

$9 W / m^{2}$

$12 W / m^{2}$

$15 W / m^{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} I_{1} = \frac{P_{1}}{4 π d_{1}^{2}} = \frac{8 π}{4 π \times 1} = 2 W / m^{2} \\ I_{2} = \frac{P_{2}}{4 π {d_{2}}^{2}} = \frac{18 π}{4 π \times (1.5)^{2}} = 2 W / m^{2} \end{array}$

Path difference

$\begin{array}{l} Δ x = d_{2} - d_{1} = 0.5 \\ = \frac{2 π}{λ} Δ x = \frac{2 π . v}{V} Δ x \\ = 2 π \times \frac{100}{300} . \frac{1}{2} = \frac{π}{3} \end{array}$