10.2 Energy Method

8.1 Kinetic Energy ($K$)
8.2 Potential Energy ($U$)
8.3 Total Energy ($E$)

10.1 Restoring Force or Torque Method
10.2 Energy Method
10.3 Important points
10.4 Concept of Pseudo Force

11.1 Different Types of Spring Block Systems

Find the stable equilibrium position, which is usually known, as the mean position. Net force or torque on the particle in this position is zero. Potential energy is minimum.

Displace the particle from its mean position by a small displacement $x$ (in case of linear SHM) or $\theta$ (in case of an angular SHM).

Find the total mechanical energy ($E$) in the displaced position. Since, mechanical energy in SHM remains constant. $$\frac{{dE}}{{dt}} = 0$$

By differentiating the energy equation with respect to time and substituting $$\frac{{dx}}{{dt}} = v,\frac{{d\theta }}{{dt}} = \omega ,\frac{{dv}}{{dt}} = a,\frac{{d\omega }}{{dt}} = \alpha $$

Convert linear acceleration and velocity into angular acceleration and velocity respectively.

Find the angular acceleration by dividing the above equation by mass $m$ and calculate the time period.

Let us derive the same expression by energy method. Suppose $\omega$ be the angular velocity of particle at angular displacement $\theta$ about point $O$.

Then, total mechanical energy of particle in position $P$ is,

$$\begin{equation} \begin{aligned} E = \frac{1}{2}I{\omega ^2} + mg\left( {{h_P} - {h_R}} \right) \\ E = \frac{1}{2}\left( {m{l^2}} \right){\omega ^2} + mgl\left( {1 - cos\theta } \right) \\ E{\text{ }}{\text{is constant}},{\text{therefore}},\frac{{dE}}{{dt}} = 0 \\ 0 = m{l^2}\omega \left( {\frac{{d\omega }}{{dt}}} \right) + mglsin\theta \left( {\frac{{d\theta }}{{dt}}} \right) \\\end{aligned} \end{equation} $$

Putting $$\frac{{d\theta }}{{dt}} = \omega ,\frac{{d\omega }}{{dt}} = \alpha {\text{ and }}sin\theta \approx \theta ,$$ we get the same expression i.e.,

$$\begin{equation} \begin{aligned} 0 = m{l^2}\omega \alpha + mgl\theta \omega \\ 0 = ml\omega \left( {l\alpha + g\theta } \right) \\\end{aligned} \end{equation} $$

Since, $$\begin{equation} \begin{aligned} ml\omega \ne 0, \\ \therefore \left( {l\alpha + g\theta } \right) = 0 \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} \alpha = - \left( {\frac{g}{l}} \right)\theta ..\left( A \right) \\ \alpha = - {\omega ^2}\theta \;\left( {{\text{SHM acceleration}}} \right)..\left( B \right) \\\end{aligned} \end{equation} $$

On equating $A$ & $B$ we get,

$$\begin{equation} \begin{aligned} \omega = \sqrt {\frac{g}{l}} ..\left( C \right) \\ T = \frac{{2\pi }}{\omega } \\ T = 2\pi \sqrt {\frac{l}{g}} \;\left( {{\text{from equation }}C} \right) \\\end{aligned} \end{equation} $$