Question

A particle executing S.H.M. has period 3 sec and amplitude 4 cm. The time it will take to travel from its position of maximum displacement to one corresponding to half its amplitude is​

Answer 1

$\bf \fbox{GIVEN} \downarrow$

$\bf {Time \:Period=3sec}$

$\bf {Amplitude\:=4cm}$

$\cos\theta=\frac{\frac{A}{2}}{A}$

$\theta=\frac{\pi}{3} = \frac{2\pi}{T}t \\ \\ t = \frac{T}{6} = \frac{3}{6} = 0.5sec$