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The displacement of a particle executing simple harmonic motion is given by $y = A_{0} + Asinωt + Bcosωt$ . Then the amplitude of its oscillation is given by

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$A + B$

$A_{0} + \sqrt{A^{2} + B^{2}}$

$\sqrt{A^{2} + B^{2}}$

$\sqrt{A_{0}^{2} + (A + B)^{2}}$

answer is C.

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Detailed Solution

$y = A_{0} + Asinωt + Bcosωt$

$or (y - A_{0}) = Asinωt + Bcosωt$

$or y^{'} = Asinωt + Bcosωt = Acos (\frac{π}{2} - ωt) + Bcosωt$

$Amplitude = \sqrt{A^{2} + B^{2} + 2 ABcos \frac{π}{2}} [∵ ϕ = \frac{π}{2}] = \sqrt{A^{2} + B^{2}}$

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