Simple harmonic motion

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Question

The position of a particle with respect to origin varies according to the relation $x = 3 \sin 100 t + 8 \cos^{2} 50 t$ .The maximum displacement of the particle from the origin is ____________.

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Solution

$\begin{array}{l} x = 3 \sin 100 t + 8 \cos^{2} 50 t \\ = 3 \sin 100 t + \frac{8 [1 + \cos 100 t]}{2} \\ x = 4 + 3 \sin 100 t + 4 \cos 100 t \\ (x - 4) = 5 \sin (100 t + ϕ) \{\tan ϕ = \frac{4}{3}\} \\ Amplitude = 5 units \\ Maximum displacement = 9 units \end{array}$

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