The time period of a simple pendulum is $T$. The time taken to complete $5/8$ oscillations starting from the mean position is $\frac{\alpha}{\beta} T$. The value of $\alpha$ is ..... .

$A$ chimpanzee swinging on a swing in a sitting position,stands up suddenly,the time period will

$A$ mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of the Earth,it is found that $t = 2T$. The entire setup is taken to the surface of another planet whose mass is half that of the Earth and radius is the same. The same experiment is repeated and corresponding times are noted as $t'$ and $T'$.

Let $l_1$ be the length of a simple pendulum. Its length changes to $l_2$ to increase the periodic time by $20 \%$. The ratio $\frac{l_2}{l_1}$ is:

$A$ simple pendulum is taken to a place where its distance from the Earth's surface is equal to the radius of the Earth. Calculate the time period of small oscillations if the length of the string is $4.0 \ m$. (Take $g = \pi^2 \ m/s^2$ at the surface of the Earth.) (in $s$)

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