Question

A particle executes SHM in a period of 8 sec. At what time after passing through its mean position will its energy be half kinetic and half potential.

Answer 1

According to question,
$$\begin{gathered} K.E=P.E \\ \Rightarrow \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2 \\ Weknow,v=\omega \sqrt{A^2-x^2} \\ weget, \\ \frac{1}{2}k{x}^2=\frac{1}{2}m(\omega \sqrt{A^2-x^2}{)}^2 \\ \Rightarrow \frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2(A^2-x^2) \\ thereforex=\frac{A}{\sqrt{2}} \end{gathered}$$
At x=$\frac{A}{\sqrt{2}}$the potential energy is equal to kinetic energy, Let t be the time to reach it at x=$\frac{A}{\sqrt{2}}$.
Here for SHM
$$\begin{gathered} x=A\sin t\left(\omega t\right) \\ forx=\frac{A}{\sqrt{2}} \\ \frac{A}{\sqrt{2}}=A\sin \left(\omega t\right),or\frac{1}{\sqrt{2}}=\sin \left(\omega t\right),or\omega t=\frac{\mathrm{\pi }}{4};t=\frac{\mathrm{\pi }}{\omega 4}=\frac{T}{8}=\frac{8}{8}=1sec \end{gathered}$$
​Hence required time is 1 sec.