Rankers Physics

Energy in SHM: Practice Problem & Solution

The bob of simple pendulum having length is displaced from mean position to an angular position $ \theta $ with respect to vertical. If it is released, then velocity of bob at lowest position: (2000)
$ \sqrt{2g(1-\cos\theta)} $
$ \sqrt{2g\ell(1+\cos\theta)} $
$ \sqrt{2g\ell\cos\theta} $
$ \sqrt{2g\ell} $

Solution Explained:

To solve this problem, we apply the core principles of Energy in SHM. Understanding the underlying formula is key to arriving at the correct answer below:

Change in potential energy equals kinetic energy at the lowest point. $ mgl(1-\cos\theta) = \frac{1}{2}mv^2 $. Solving for velocity gives $ v = \sqrt{2gl(1-\cos\theta)} $. (Note: length parameter $ l $ is implied in option a despite typo).

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