A particle in SHM has displacement $x = a\cos(\omega t) + b\sin(\omega t)$. The amplitude of motion is:
A$|a| + |b|$
B$\sqrt{a^2 + b^2}$
C$\sqrt{a^2 - b^2}$
D$\max(|a|,|b|)$
Answer & Solution
Correct answer: B. $\sqrt{a^2 + b^2}$
Express as single sinusoid: $x = R\sin(\omega t + \phi)$ with $R\sin\phi = a$ and $R\cos\phi = b$. So $R^2 = a^2 + b^2$ ⇒ $R = \sqrt{a^2+b^2}$. (Classic Asin + Bcos identity.)
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