Derive the equation for the kinetic energy and potential energy of a simple harmonic oscillator and show that the total energy of a particle in simple harmonic motion is constant at any point on its path.

Expression for Kinetic energy: Consider any body or system executing SHM is called simple harmonic oscillator particle of mass m executing SHM between extreme positions x = + A and x = –A.
Displacement of the particle in SHM when it starts from extreme position is given by
$\mathrm{x}=\mathrm{Acos}\mathrm{\omega t}$
Velociy of the particle executing SHM is given by $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{Acos}\mathrm{\omega t})}{\mathrm{dt}}$

$\begin{array}{l}=-\mathrm{A\omega sin}\mathrm{\omega t}=-\mathrm{A\omega }\sqrt{1-{\cos }^2\mathrm{\omega t}}\\ \therefore \mathrm{V}=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2}[\because \mathrm{x}=\mathrm{acos}\mathrm{\omega t}]\end{array}$

Therefore, KE of the particle in SHM is given by $\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}{\mathrm{mA}}^2{\mathrm{\omega }}^2{\mathrm{Sin}}^2\mathrm{\omega t}$

$=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}^2\right)$

Expression for Potential Energy: When a particle executing SHM is displaced through ‘x’ from equilibrium position, then restoring force acting on it is F = –kx

Further, if it displaces through ‘dx’ then work done against the restoring force is given by

$\mathrm{dw}=-\mathrm{Fdx}=-(-\mathrm{kxdx})=\mathrm{kxdx}$

Therefore, work done against the restoring force in moving the particle from equilibrium position (x=0) to any displacement (x) is given by

$\begin{array}{l}\mathrm{W}=\int \mathrm{dw}={\int }_0^{\mathrm{x}} \mathrm{kxdx}=\mathrm{k}{\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^{\mathrm{x}}=\frac{1}{2}{\mathrm{kx}}^2=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2\\ \left(\because {\mathrm{\omega }}^2=\frac{\mathrm{k}}{\mathrm{m}}\right)\end{array}$

This work done against the restoring force is stored in the bob as its potential energy.
$\begin{array}{l}\therefore \mathrm{PE}=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2{\cos }^2\mathrm{\omega t}\\ (\because \mathrm{x}=\mathrm{Acos}\mathrm{\omega t})\end{array}$

Conservation of total energy of Simple Harmonic Oscillator :

At mean position or at x = 0 :

$\begin{array}{l}\therefore \mathrm{KE}=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}^2\right)=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-0^2\right)\\ =\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2\\ \therefore \mathrm{PE}=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2=\frac{1}{2}{\mathrm{m\omega }}^2(0{)}^2=0\end{array}$

Total Energy

$\mathrm{E}=\mathrm{KE}+\mathrm{PE}=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2+0=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2\to \left(1\right)$

$\text{ At extreme position or at }\mathrm{x}=\pm \mathrm{A}$

$\begin{array}{l}\mathrm{KE}=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}^2\right)=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-{\mathrm{A}}^2\right)=0\\ \mathrm{PE}=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2\\ \therefore \text{ Total Energy }\\ \mathrm{E}=\mathrm{KE}+\mathrm{PE}=0+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2\to \left(2\right)\end{array}$

At any position(x):
$\mathrm{KE}=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}^2\right)\text{ and }\mathrm{PE}=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2$

Total Energy
$\begin{array}{l}\mathrm{E}=\mathrm{KE}+\mathrm{PE}=\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}^2\right)+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2\\ =\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{A}}^2\to \left(3\right)\end{array}$

Hence from equation (1), (2), and (3) total energy of a simple harmonic oscillator is constant at any point on this path.

The variation of KE and PE with displacement (x) is shown in the following graph