$A$ $U$-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 of mass $0.01 \ kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force $F$ as shown in the graph (where $F$ is in $N$ and $x$ is in $m$). The period of the $S.H.M.$ is ... $s$. (Given: The graph is a straight line passing through the origin with slope $k = 4 \ N/m$ calculated from $F = 8 \ N$ at $x = 2 \ m$).

$9 \text{ kg}$ 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).

Under the influence of force $F_1$,a body oscillates with a period $T_1$,and due to another force $F_2$,the body oscillates with a period $T_2$. If both forces act simultaneously,what is the resultant period? (Consider the displacement is the same in all three cases.)

$A$ particle is placed at the lowest point of a smooth wire frame in the shape of a parabola,lying in the vertical $xy-$ plane having equation $x^2 = 5y$ ($x, y$ are in meters). After a slight displacement,the particle is set free. Find the angular frequency of oscillation in $rad/s$ (take $g = 10 \ m/s^2$).

$A$ 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: