The displacement of a particle executing simple harmonic motion is given by $y = A_0 + A\sin\omega t + B\cos\o

Question

The displacement of a particle executing simple harmonic motion is given by $y = A_0 + A\sin\omega t + B\cos\omega t$. Then the amplitude of its oscillation is given by :

Accepted Answer

Given $y = A_0 + A\sin\omega t + B\cos\omega t$. Equate SHM: $y' = y - A_0 = A\sin\omega t + B\cos\omega t$. Resultant amplitude $R = \sqrt{A^2 + B^2 + 2AB\cos 90^\circ} = \sqrt{A^2 + B^2}$.

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