The motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is

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simple harmonic with amplitude $\sqrt$(A^2 + B^2) x = A sin ωt + B cos ωt ...(1) \text { Acceleration }, a = \frac{\text {d}^2 x}{\text {dt}^2} = \frac{\text {d}^2}{\text{dt}^2}(\text {A}\sin\omega t + \text {B} \cos \omega t) = \frac{\text{d}}{\text {dt}}(\text { A }\omega \cos \omega t - \text { B }\omega \sin \omega t) = - \text { A } \omega^2 \text { sin }\omega t - \text { B }\omega^2 \cos \omega t = - \omega^2 (\text { A }\sin \omega t + \text { B }\cos \omega t ) = - \omega^2 x For a body to undergo simple harmonic motion, acceleration, a =- kx. ...(2) Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude, A = $\sqrt$(A^2 + B^2)

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