Question

A simple pendulum of length $l$ is made to oscillate with amplitude of $45$ degrees. The acceleration due to gravity is $g$. Let $T_0=2\pi \sqrt{l/g}$. The time period of oscillation of this pendulum will be.

• A. $T_0$ irrespective of the amplitude
• B. Slightly less than $T_0$
• C. Slightly more then $T_0$
• D. Dependent on whether it swings in a plane aligned with the north-south or east west directions

Answer 1

The correct option is C Slightly more then $T_0$

As the $T_0=2\pi \sqrt{\frac{l}{g}}$ is defined for very small amplitude($\theta$) i.e., $\theta <<1$

Actually, $T=2\pi \sqrt{\frac{l}{g}}(1+\frac{{\theta }_0^2}{16}+...)$

So with increase in amplitude, the period also increases gradually.

Hence time period of given oscillation with amplitude 45 degrees will be slightly more than $T_0$.