At what displacement is the kinetic energy of a particle performing simple harmonic motion of amplitude 10 cm is three times it’s potential energy?

KE=3(PE)

$$\begin{gathered} we know,Kinetic energy KE, Potential energy PE as \\ \text{KE=}\frac{1}{2}{\text{mω}}^{\text{2}}\left({\text{A}}^{\text{2}}{\text{-x}}^{\text{2}}\right) \\ PE\text{=}\frac{1}{2}{\text{mω}}^{\text{2}}\left({\text{x}}^{\text{2}}\right) \\ by the question, KE = 3 PE \\ \Rightarrow \frac{1}{2}{\text{mω}}^{\text{2}}\left({\text{A}}^{\text{2}}{\text{-x}}^{\text{2}}\right)\text{=}\frac{3}{2}{\text{mω}}^{\text{2}}\left({\text{x}}^{\text{2}}\right) \\ on simplification, {\text{A}}^{\text{2}}{\text{=4x}}^{\text{2}} \\ \Rightarrow \text{x=A/2} \\ Given A=10 cm \\ \therefore \text{x}=\frac{10}{2}=\text{5cm} \end{gathered}$$