Physics - SHM Question with Solution | TestHub

$\begin{array}{rl}& \text{ Given : K.E }=\text{ P.E }+\frac{25}{100}.P.E\\ & \mathrm{K}.\mathrm{E}=\mathrm{P}.\mathrm{E}+\frac{1}{4}\mathrm{P}.\mathrm{E}\\ & \text{ We know }\mathrm{K}.\mathrm{E}=\frac{1}{2}\mathrm{m}{\omega }^2\left({\mathrm{A}}^2-{\mathrm{x}}^2\right)\text{ and }\mathrm{P}\cdot \mathrm{E}=\frac{1}{2}\mathrm{m}{\omega }^2{\mathrm{x}}^2\\ & \therefore \mathrm{K}.\mathrm{E}=\frac{5}{4}\text{ P.E }\\ & \frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{5}{4}\left(\frac{1}{2}m{\omega }^2{x}^2\right)\\ & {\mathrm{A}}^2-{\mathrm{x}}^2=\frac{5}{4}{\mathrm{x}}^2\\ & {\mathrm{A}}^2=\frac{9}{4}{\mathrm{x}}^2\\ & \therefore \mathrm{A}=\frac{3}{2}\mathrm{x}\\ & \therefore \mathrm{x}=\mathrm{A}\times \frac{2}{3}=2\mathrm{cm}\\ & \end{array}$