A simple pendulum has a time period of $T=4\text{ s}$ on Earth. If it is taken to a planet where the acceleration due to gravity is $g^{\prime }=\frac{g}{4}$, what will be its new time period $T^{\prime }$?

Solution

1. The time period of a simple pendulum is given by the formula $T=2\pi \sqrt{\frac{L}{g}}$.
2. Given that the new gravity is $g^{\prime }=\frac{g}{4}$, the new time period is $$\begin{array}{rl}{T}^{\prime }& =2\pi \sqrt{\frac{L}{g^{\prime }}}\\ & =2\pi \sqrt{\frac{L}{g/4}}\end{array}$$.
3. Simplifying the expression gives $$\begin{array}{rl}{T}^{\prime }& =2\pi \sqrt{\frac{4L}{g}}\\ & =2\times (2\pi \sqrt{\frac{L}{g}})\\ & =2T\end{array}$$.
4. Substituting the given value $T=4\text{ s}$, we find $$\begin{array}{rl}{T}^{\prime }& =2\times 4\text{ s}\\ & =8\text{ s}\end{array}$$.
5. Hence the answer is (C).