Physics - SHM Question with Solution | TestHub

Lets take simplest equation of SHM, $y=A\sin (\omega t)$
Given, $A=6\mathrm{cm},\mathrm{T}=3\sec$ $\therefore \omega =\frac{2\pi }{T}=\frac{2\pi }{3}{\sec }^{-1}.$ Time when particle reaches extreme position $\mathrm{P}$ is $1.5\sec$. Let time to reach $\mathrm{Q}$ is $\delta$. $\begin{array}{rl}& \Rightarrow \mathrm{y}=3\mathrm{cm}=6\mathrm{cm}\sin \left(\frac{\pi }{2}+\omega \delta \right)\\ & \Rightarrow \sin \left(\frac{\pi }{2}+\omega \delta \right)=\frac{1}{2}\\ & \Rightarrow \cos (\omega \delta )=\frac{1}{2}\\ & \Rightarrow \delta =\frac{\pi }{3\omega }=\frac{\pi (3)}{3(2\pi )}=0.5\mathrm{s}\end{array}$