17.5 Energy in SHM

17.5 Energy in Simple Harmonic Motion

Interchange Between Kinetic and Potential Energy

Conservation of Energy

• In an undamped SHM system, energy continuously transfers between Kinetic Energy ($E_k$) and Potential Energy ($E_p$).
• Total Energy $E_T$ remains constant.

Energy in a Pendulum

Kinetic Energy

Derivation: Kinetic Energy in SHM

The kinetic energy of an oscillator is given by the standard equation: $E_k=\frac{1}{2}m{v}^2$

1. Kinetic Energy in terms of displacement ($x$) Substituting the SHM velocity-displacement equation $v=\pm \omega \sqrt{x_0^2-x^2}$:

$$\begin{array}{rl}{E}_k& =\frac{1}{2}m{\left(\pm \omega \sqrt{x_0^2-x^2}\right)}^2\\ E_k& =\frac{1}{2}m{\omega }^2(x_0^2-x^2)\end{array}$$

2. Kinetic Energy in terms of time ($t$) Alternatively, substituting the SHM velocity-time equation $v=-x_0\omega \sin (\omega t)$:

$$\begin{array}{rl}{E}_k& =\frac{1}{2}m(-x_0\omega \sin (\omega t){)}^2\\ E_k& =\frac{1}{2}m{\omega }^2{x}_0^2{\sin }^2(\omega t)\end{array}$$

3. Maximum Kinetic Energy Kinetic energy reaches its maximum at the equilibrium position, where displacement $x=0$. Substituting $x=0$ into our first derived equation:

$$\begin{array}{rl}{E}_{k(\text{max})}& =\frac{1}{2}m{\omega }^2(x_0^2-0^2)\\ E_{k(\text{max})}& =\frac{1}{2}m{\omega }^2{x}_0^2\end{array}$$

Total Energy

• Because Potential Energy is zero at equilibrium, the maximum kinetic energy is equal to the Total Energy of the system.

Therefore, the Total Energy ($E_T$) of the system is given by: $E_T=\frac{1}{2}m{\omega }^2{x}_0^2$