Physics - SHM Question with Solution | TestHub

The total energy is given by T.E $=\frac{1}{2}{\mathrm{kA}}^2$ Kinetic energy is given by K.E $=\frac{1}{2}k\left(A^2-x^2\right)$ $$\begin{array}{ll}& \text{ K.E }=\frac{1}{2}P.E\\ \therefore & \frac{1}{2}k\left(A^2-x^2\right)=\frac{1}{2}\left(\frac{1}{2}{\mathrm{kA}}^2\right)\\ \therefore & \left({\mathrm{A}}^2-{\mathrm{x}}^2\right)=\frac{1}{2}{\mathrm{A}}^2\\ \therefore & \left({\mathrm{A}}^2-{\mathrm{x}}^2\right)=\frac{1}{2}{\mathrm{A}}^2\\ \therefore & \mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}}\end{array}$$ The equation of displacement in SHM is $$\begin{array}{ll}& \mathrm{x}=\mathrm{A}\sin \frac{2\pi \mathrm{t}}{\mathrm{T}}\\ \therefore & \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A}\sin \frac{2\pi \mathrm{t}}{4}\\ \therefore \frac{1}{\sqrt{2}}=\sin \frac{\pi \mathrm{t}}{2}& \dots .(\because \mathrm{T}=4\sec )\\ \therefore \frac{\pi \mathrm{t}}{2}& ={\sin }^{-1}\frac{1}{\sqrt{2}}\\ & \frac{\pi \mathrm{t}}{2}=\frac{\pi }{4}\\ \therefore & \mathrm{t}=0.5\mathrm{s}\end{array}$$