$A$ 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?

$A$ cone with half the density of water is floating in water as shown in the figure. It is depressed down by a small distance $\delta (\ll H)$ and released. The frequency of simple harmonic oscillations of the cone 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 executes linear simple harmonic motion with an amplitude of $2 \ cm$. When the particle is at $1 \ cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod $AB$ of length $l$ and mass $m$. The rod is pivoted at its centre $O$ and can rotate freely in a horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in the figure. The rod is gently pushed through a small angle $\theta$ and released. The frequency of the resulting oscillation is

$9 \ kg$ of mercury is poured into a glass $U$-tube with an inner diameter of $1.2 \ cm$. The mercury can flow without friction within the tube. The oscillation period is ......... $\sec$. (Density of mercury $\rho = 13.6 \times 10^3 \ kg/m^3$)