Physics - SHM Question with Solution | TestHub

Potential energy of a simple harmonic oscillator $$U=\frac{1}{2}m{\omega }^2{y}^2$$ Kinetic energy of a simple harmonic oscillator $$K=\frac{1}{2}m{\omega }^2\left(A^2-y^2\right)$$ Here $y=$ displacement from mean position $A=$ maximum displacement (or amplitude) from mean position Total energy is $$\begin{array}{rl}E& =U+K\\ & =\frac{1}{2}m{\omega }^2{y}^2+\frac{1}{2}m{\omega }^2\left(A^2-y^2\right)\\ & =\frac{1}{2}m{\omega }^2{A}^2\end{array}$$ When the particle is half way to its end point ie, at half of its amplitude then $$y=\frac{A}{2}$$ Hence, potential energy $$\begin{array}{rl}U& =\frac{1}{2}m{\omega }^2{\left(\frac{A}{2}\right)}^2\\ & =\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)\\ U& =\frac{E}{4}\end{array}$$