If $x_1 = A\sin(\omega t)$ and $x_2 = A\sin(\omega t + \pi/3)$, the resultant amplitude of $x_1 + x_2$ is:
A$A$
B$\sqrt 3 A$
C$\sqrt 2 A$
D$2A$
Answer & Solution
Correct answer: B. $\sqrt 3 A$
Two SHMs of same $\omega$ add as vectors (phasors). Resultant amplitude: $A_R = \sqrt{A^2 + A^2 + 2A^2\cos(\pi/3)} = \sqrt{2A^2(1+\cos 60°)} = \sqrt{2A^2 \cdot 3/2} = \sqrt{3}A$.
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