8.1 Kinetic Energy ($K$)

  Simple Harmonic Motion

Let us consider the equation of SHM, $${x} = {\bf{A}}\sin \left( {{\bf{\omega t}} + \phi } \right)$$

Differentiating it w.r.t. time we get, $$\frac{{dx}}{{dt}} = v = A\omega \cos \left( {\omega t + \phi } \right)$$ So, the kinetic energy of the particle is,

$$\begin{equation} \begin{aligned} K = \frac{1}{2}m{v^2} = \frac{1}{2}m{\left[ {A\omega \cos \left( {\omega t + \phi } \right)} \right]^2} \\ K = \frac{1}{2}m\left[ {{A^2}{\omega ^2} - {A^2}{\omega ^2}si{n^2}\left( {\omega t + \phi } \right)} \right] \\ K = \frac{1}{2}m\left[ {{A^2}{\omega ^2} - {\omega ^2}{x^2}} \right] \\ K = \frac{1}{2}m{\omega ^2}\left[ {{A^2} - {x^2}} \right] \\ K = \frac{1}{2}k\left[ {{A^2} - {x^2}} \right]\quad \left( {{\text{where }}k = m{\omega ^2}} \right) \\\end{aligned} \end{equation} $$

From this expression we can see that the kinetic energy is maximum at the center and zero at the extremes of oscillation $\left( {x = \pm A} \right)$.