AIPMT PRELIMS 2000 Physics Simple Pendulum MCQ Question

When the bob of a simple pendulum is displaced to an angle $\theta$ and released, it converts its potential energy at the height back into kinetic energy at the lowest point. The potential energy at the height can be expressed as $PE = mgh$, where $h = \ell(1 - \cos \theta)$ is the height above the lowest point. Thus, $PE = mg\ell(1 - \cos \theta)$. At the lowest point, all this potential energy is converted into kinetic energy, given by $KE = \frac{1}{2} mv^2$. Setting these equal gives us: $mg\ell(1 - \cos \theta) = \frac{1}{2} mv^2$ Cancelling $m$ and rearranging, we find: $v^2 = 2g\ell(1 - \cos \theta)$ Taking the square root, we obtain: $v = \sqrt{2g\ell(1 - \cos \theta)}$ Thus, the correct answer is A.

The other options are incorrect because:

• Option B: $\sqrt{2g\ell(1 + \cos \theta)}$ does not correctly represent the energy conversion.
• Option C: $\sqrt{2g\ell \cos \theta}$ is also incorrect as it does not account for the height change.
• Option D: $\sqrt{2g\ell}$ ignores the angle $\theta$ and the corresponding height change.