Question

Light of energy E incident on bob of simple pendulum, Find its amplitude

Answer 1

• Exact amplitude:
  $\displaystyle \theta = \cos^{-1}\left(1-\frac{E}{mgL}\right)$

• For small angles:
  $\displaystyle \theta \approx \sqrt{\frac{2E}{mgL}}$

Answer 2

The energy $E$ absorbed by the bob gets converted into gravitational potential energy at the maximum displacement. For a pendulum of length $L$ and mass $m$, if the maximum angular displacement is $\theta$, the vertical height risen is

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

Thus, by energy conservation,

$$E = mgh = mgL(1-\cos\theta).$$

This gives:

$$\cos\theta = 1 - \frac{E}{mgL}.$$

So, the exact amplitude (angular) is

$$\theta = \cos^{-1}\left(1-\frac{E}{mgL}\right).$$

For small amplitudes, we use the small angle approximation:

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

Then,

$$1 - \frac{\theta^2}{2} \approx 1 - \frac{E}{mgL} \quad \Longrightarrow \quad \frac{\theta^2}{2} \approx \frac{E}{mgL}.$$

Thus,

$$\theta \approx \sqrt{\frac{2E}{mgL}}.$$

Explanation (minimal):

1. Equate absorbed energy $E$ to gravitational potential energy $mgL(1-\cos\theta)$.
2. Solve for $\theta$ exactly: $\theta = \cos^{-1}\left(1-\frac{E}{mgL}\right)$.
3. Use small angle approximation to get $\theta \approx \sqrt{\frac{2E}{mgL}}$.