A simple pendulum has a length of $L=1\text{ m}$. What is its time period of oscillation in seconds? Take $g=10{\text{ m/s}}^2$ and ${\pi }^2=10$.

Solution

1. The formula for the time period of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$.
2. Substituting $L=1$ and $g=10$, we get $T=2\pi \sqrt{\frac{1}{10}}$.
3. Since ${\pi }^2=10$, then $\pi =\sqrt{10}$, so $$\begin{array}{rl}T& =2\sqrt{10}\times \frac{1}{\sqrt{10}}\\ & =2\text{ s}\end{array}$$.
4. Hence the answer is (B).