$A$ uniform rod of mass $m$ and length $l$ is suspended about its end. The time period of small angular oscillations is ..........

$A$ rectangular block of mass $m$ and area of cross-section $A$ floats in a liquid of density $\rho$. If it is given a small vertical displacement from equilibrium,it undergoes simple harmonic motion with a time period $T$. Then:

$A$ particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$. Then,its time period of vibration will be

$A$ $U$-tube of uniform bore of cross-sectional area '$A$' is set up vertically. '$M$' grams of a liquid of density '$d$' is poured into it. The column of liquid in this tube will oscillate with a period '$T$', which is equal to [$g$ = acceleration due to gravity]

$A$ body executes simple harmonic motion under the action of a force $F_1$ with a time period $T_1 = \frac{4}{5} \ s$. If the force is changed to $F_2$, it executes $SHM$ with a time period $T_2 = \frac{3}{5} \ s$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body, what will be its new time period in seconds?

$P$ and $Q$ are fixed points in the same plane and a mass $m$ is tied by strings as shown in the figure. If the mass is displaced slightly out of this plane and released,it will oscillate with a time period (given $PQ = 2d$,$PR = QR = L$).