$A$ body performs $S.H.M.$ under the action of force $F_1$ with period $T_1$ seconds. If the force is changed to $F_2$,it performs $S.H.M.$ with period $T_2$ seconds. If both forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,the period in seconds will be

$A$ $U$-tube of uniform bore of cross-sectional area $A$ has been set up vertically with open ends facing up. Now $M$ grams of a liquid of density $d$ is poured into it. The column of liquid in this tube will oscillate with a period $T$ such that

$A$ particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x) = \frac{1}{2} k x^2 - V_0 \cos \left(\frac{x}{a}\right)$,where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$,the time period is given by

$9 \text{ kg}$ of solution is poured into a glass $U$-tube as shown in the figure. The tube's inner diameter is $2 \sqrt{\frac{\pi}{5}} \text{ m}$ and the solution oscillates freely up and down about its position of equilibrium $(x=0)$. The period of oscillation in seconds is (Given: $1 \text{ m}^3$ of solution has a mass $\rho=900 \text{ kg/m}^3$, $g=10 \text{ m/s}^2$, ignore frictional and surface tension effects).

$A$ body executes simple harmonic motion under the action of a force $F_1$ with a time period $(4/5) \ s$. If the force is changed to $F_2$,it executes $SHM$ with a time period $(3/5) \ s$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,its time period (in $s$) is:

$A$ solid cube of side $l$ is made to oscillate about a horizontal axis passing through one of its edges. Its time period will be