If $y_1 = 5 (mm)\ \sin\pi t$ is equation of oscillation of source $S_1$ and $y_2$ $=$ $5$ $(mm)$ $sin(\pi t + \pi /6)$ be that of $S_2$ and it takes $1$ $sec$ and $\frac{1}{2}\ sec$ for the transverse waves to reach point $A$ from sources $S_1$ and $S_2$ respectively then the resulting amplitude at point $A$, is .... $mm$
Diffcult
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Wave originating at $t=0$ from $S_{1}$ reaches point $A$ at $t=1$
Wave originating at $t=\frac{1}{2}$ from $S_{2}$ reaches point $A$ at $t=1$
So phase difference in these waves
$=\frac{\pi}{2}+\frac{\pi}{6}, A=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi}=5$
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