Question

The displacement of particle performing S.H.M. is given by
$x=[5sin\pi t+12sin(\pi t+\frac{\pi }{2})]$ cm. Determine the amplitude, period and initial phase of the motion.

Answer 1

$x=5\sin \pi t+12\sin (\pi t+\pi /2)$

$=5\sin \pi t+12\cos \pi t$

$=13\sin (\pi t+\theta )$ where $\cos \theta =5/12$

Amplitude, $A=max\left(x\right)=13cm$

Time period, $T=2\pi /\omega =2\pi /\pi =2s$

Initial phase, $\alpha =\angle (x{)}_{t=0}=\theta$

$\alpha ={\cos }^{-1}\left(5/12\right)$