Question

A simple pendulum of length 'L' has mass 'M' and it oscillates freely with amplitude 'A'. At extreme position, its potential energy is

• A. $\frac{MgA^2}{L}$
• B. $\frac{2MgA^2}{L}$
• C. $\frac{MgA}{2L}$
• D. $\frac{MgA^2}{2L}$

Answer 1

Answer: (D)

$\frac{MgA^2}{2L}$

Answer 2

For a simple pendulum, the vertical height increase at an angular displacement $\theta$ is given by

$$h = L(1 - \cos\theta).$$

Given that the amplitude in arc length is $A = L\theta$, for small angles we use the approximation

$$\cos\theta \approx 1 - \frac{\theta^2}{2}.$$

Thus,

$$h \approx L\left(1 - \left(1 - \frac{\theta^2}{2}\right)\right) = \frac{L\theta^2}{2} = \frac{L}{2}\left(\frac{A}{L}\right)^2 = \frac{A^2}{2L}.$$

The potential energy at the extreme position is

$$PE = Mgh = Mg\left(\frac{A^2}{2L}\right) = \frac{MgA^2}{2L}.$$