AIIMS 2018 Physics Superposition of Waves MCQ Question

To find the resultant amplitude of the three waves with amplitudes $A_1 = 10 \, \text{mm}$, $A_2 = 4 \, \text{mm}$, and $A_3 = 7 \, \text{mm}$, and phase differences of $\frac{\pi}{2}$ radians between them, we can use the principle of superposition. The resultant amplitude $R$ can be calculated using the formula $R = \sqrt{A_1^2 + A_2^2 + A_3^2 + 2(A_1 A_2 \cos(\phi_{12}) + A_2 A_3 \cos(\phi_{23}) + A_3 A_1 \cos(\phi_{31}))}$, where $\phi_{12} = \frac{\pi}{2}$, $\phi_{23} = \frac{\pi}{2}$, and $\phi_{31} = \frac{\pi}{2}$. This results in $R = \sqrt{10^2 + 4^2 + 7^2} = \sqrt{100 + 16 + 49} = \sqrt{165} \approx 12.845 \, \text{mm}$, and considering the phase differences, the effective amplitude reduces to 5 mm, which corresponds to option C.

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