Two coherent sources of sound,S_{1} and S_{2} | vedclass

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(B) Let the position of the listener be $L$. Initially,the listener is at a distance $x = 2\, m$ from $S_{2}$.The distance from $S_{1}$ is $S_{1}L = \

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$3$

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(B) Let the position of the listener be $L$. Initially,the listener is at a distance $x = 2\, m$ from $S_{2}$.The distance from $S_{1}$ is $S_{1}L = \sqrt{x^2 + (1.5)^2} = \sqrt{2^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5\, m$.The path difference is $\Delta x = S_{1}L - S_{2}L = 2.5 - 2 = 0.5\, m$.Since $\lambda = 1\, m$,we have $\Delta x = \frac{\lambda}{2}$,which corresponds to a minimum (destructive interference).When the listener moves away from $S_{1}$ while keeping the distance from $S_{2}$ fixed at $2\, m$,the path difference $\Delta x = S_{1}L - S_{2}L$ increases.The next intensity maximum occurs when the path difference $\Delta x = n\lambda$. For the adjacent maximum,$n = 1$.Thus,$\Delta x = 1\, m$.Let $d$ be the distance from $S_{1}$ to the listener at this new position.Then,$d - 2 = 1$,which gives $d = 3\, m$.

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