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  4. A particle performs S.H.M. with amplitude 25 cm and period 3 s. The minimum time required for it to move between two points 12.5 cm on either side of the mean position is

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Q. A particle performs $S$.$H$.$M$. with amplitude $25\,cm$ and period $3 \,s$. The minimum time required for it to move between two points $12.5\, cm$ on either side of the mean position is

MHT CETMHT CET 2015

Solution:

$ \sin \theta=\frac{12.5}{25} $
$ \theta=\frac{\pi}{6} $
So, total angle to move $=\frac{2 \pi}{6}=\frac{\pi}{3}$
It takes 3 s to rotate $360^{\circ}$. So, to rotate 60
$ t=\frac{3}{360} \times 60 $
$ t =0.5 s $