In an experiment for determining the gravitational acceleration $g$ of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure. What is the value of $g$ at the place? ...... $m/s^2$

If two persons sitting on a swing instead of one, why the periodic time does not changed ?

A simple pendulum of length $l$ is made to oscillate with an 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

The graph in figure represents

A simple pendulum of length $L$ and mass (bob) $M$ is oscillating in a plane about a vertical line between angular limits $ - \varphi $ and $ + \varphi $. For an angular displacement $\theta (|\theta | < \varphi )$, the tension in the string and the velocity of the bob are $T$ and $ v$ respectively. The following relations hold good under the above conditions

The acceleration due to gravity at a place is ${\pi ^2}\,m/se{c^2}$. Then the time period of a simple pendulum of length one metre is