If y_1 = 5 text{ mm} sin(pi t) is the equatio | vedclass

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(C) The displacement of the wave from $S_1$ at point $A$ at time $t$ is $y_{A1} = 5 \sin(\pi(t - 1))$.The displacement of the wave from $S_2$ at point

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$5$

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(C) The displacement of the wave from $S_1$ at point $A$ at time $t$ is $y_{A1} = 5 \sin(\pi(t - 1))$.The displacement of the wave from $S_2$ at point $A$ at time $t$ is $y_{A2} = 5 \sin(\pi(t - 0.5) + \pi/6)$.Comparing the phases at point $A$:Phase of wave $1$: $\phi_1 = \pi t - \pi$.Phase of wave $2$: $\phi_2 = \pi t - 0.5\pi + \pi/6 = \pi t - \pi/2 + \pi/6 = \pi t - \pi/3$.The phase difference $\Delta\phi = \phi_2 - \phi_1 = (\pi t - \pi/3) - (\pi t - \pi) = \pi - \pi/3 = 2\pi/3$.The resulting amplitude $A_{res} = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\Delta\phi)}$.Given $A_1 = A_2 = 5 	ext{ mm}$ and $\Delta\phi = 2\pi/3$:$A_{res} = \sqrt{5^2 + 5^2 + 2(5)(5) \cos(2\pi/3)} = \sqrt{25 + 25 + 50(-0.5)} = \sqrt{50 - 25} = \sqrt{25} = 5 	ext{ mm}$.

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