Physics - SHM Question with Solution | TestHub

Particle executing SHM. Displacement $y=5\sin \left(4t+\frac{\pi }{3}\right)...(i)$ Velocity of particle $$\begin{array}{l}\left(\frac{dy}{dt}\right)=\frac{5d}{dt}\sin \left(4t+\frac{\pi }{3}\right)\\ =5\cos \left(4t+\frac{\pi }{3}\right)\cdot 4=20\cos \left(4t+\frac{\pi }{3}\right)\end{array}$$ Velocity at $t=\left(\frac{T}{4}\right)$ $${\left(\frac{dy}{dt}\right)}_{t=\frac{T}{4}}=20\cos \left(4\times \frac{T}{4}+\frac{\pi }{3}\right)$$ or $u=20\cos \left(T+\frac{\pi }{3}\right)...(ii)$ Comparing the given equation with standard equation of SHM. $y=a\sin (\omega t+\varphi )$ We get, $\omega =4$ As $\omega =\frac{2\pi }{T}\Rightarrow T=\frac{2\pi }{\omega }$ or $T=\frac{2\pi }{4}=\left(\frac{\pi }{2}\right)$ Now, putting value of $T$ in Eq. (ii), we get $$\begin{array}{rl}u& =20\cos \left(\frac{\pi }{2}+\frac{\pi }{3}\right)=-20\sin \frac{\pi }{3}\\ & =-10\times \sqrt{3}\end{array}$$ The kinetic energy of particle, $$\begin{array}{rl}\mathrm{KE}& =\frac{1}{2}m{u}^2=\frac{1}{2}\times 2\times 10^{-3}\times (-10\sqrt{3}{)}^2\\ & =10^{-3}\times 100\times 3=0.3\mathrm{J}\end{array}$$