$A$ 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

$A$ drinking straw is dipped in a pan of water to a depth $d$ from the surface (see figure). Water is sucked into it up to an initial height $h_0$ and then left to oscillate. As a result,its height $y$ from the surface of the water varies periodically. Ignoring damping,the equation for $y$ is ($g$ is the acceleration due to gravity):

$A$ large horizontal surface moves up and down in $S.H.M.$ with an amplitude of $1 \, cm$. If a mass of $10 \, kg$ placed on the surface is to remain continuously in contact with it,the maximum frequency of $S.H.M.$ will be .... $Hz$.

$A$ 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?

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

$A$ particle executes linear $S.H.M.$ with amplitude $4 \ cm$. The magnitude of velocity and acceleration is equal when it is at $3 \ cm$ from the mean position. The time period of oscillation is: