Two pendulums have time periods $T$ and $5T/4$. They start $SHM$ at the same time from the mean position. After how many oscillations of the smaller pendulum will they be again in the same phase?

The motion of a simple pendulum executing $S.H.M.$ is represented by the following equation:
$Y = A \sin (\pi t + \phi)$,where time is measured in $seconds$.
The length of the pendulum is .............$cm$.

$A$ pendulum is suspended by a string of length $250\,cm$. The mass of the bob of the pendulum is $200\,g$. The bob is pulled aside until the string is at $60^{\circ}$ with the vertical,as shown in the figure. After releasing the bob,the maximum velocity attained by the bob will be . . . . . . $m/s$. (if $g = 10\,m/s^2$)

In an experiment to determine the period of a simple pendulum of length $1\, m$,it is attached to different spherical bobs of radii $r_1$ and $r_2$. The two spherical bobs have uniform mass distribution. If the relative difference in the periods is found to be $5 \times 10^{-4}$,the difference in radii,$|r_1 - r_2|$,is best given by .... $cm$.

The metallic bob of a simple pendulum has a relative density of $5$. The time period of this pendulum is $10 \, s$. If the metallic bob is immersed in water,then the new time period becomes $5 \sqrt{x} \, s$. The value of $x$ will be.

If the length of a seconds pendulum becomes $\frac{1}{3}$ of its original length,what will be its new periodic time?