Two loudspeakers (L_1 and L_2) are placed wit | vedclass

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(B) Let the distance of point $A$ from the line joining the speakers be $D = 40 \ m$. The speakers $L_1$ and $L_2$ are separated by $10 \ m$, so their

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$600$

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(B) Let the distance of point $A$ from the line joining the speakers be $D = 40 \ m$. The speakers $L_1$ and $L_2$ are separated by $10 \ m$, so their coordinates relative to the midpoint of the line joining them are $(0, 5)$ and $(0, -5)$. Point $A$ is at $(40, 0)$.When the recorder moves to point $B$ at a distance of $25 \ m$ from $A$, its coordinates are $(40, 25)$.The distances from the speakers to point $B$ are:$L_1B = \sqrt{40^2 + (25-5)^2} = \sqrt{40^2 + 20^2} = \sqrt{1600 + 400} = \sqrt{2000} = 20\sqrt{5} \ m$.Given $\sqrt{5} = 2.23$, $L_1B = 20 	imes 2.23 = 44.6 \ m$.$L_2B = \sqrt{40^2 + (25+5)^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \ m$.The path difference at point $B$ is $\Delta x = L_2B - L_1B = 50 - 44.6 = 5.4 \ m$.Since the recorder undergoes $10$ cycles of minima and maxima, point $B$ corresponds to the $10^{	ext{th}}$ maxima, so $\Delta x = n\lambda$, where $n = 10$.$5.4 = 10 	imes \lambda \implies \lambda = 0.54 \ m$.The frequency $f$ is given by $f = \frac{v}{\lambda} = \frac{324}{0.54} = 600 \ Hz$.

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