$A$ particle of mass $m$ is located in a one-dimensional potential field where the potential energy is given by: $V(x) = A(1 - \cos px)$,where $A$ and $p$ are constants. The period of small oscillations of the particle is

$A$ 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?

$A$ 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:

$A$ $U$-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

The potential energy of a particle of mass $10 \ g$ as a function of displacement $x$ is $(50 x^2 + 100) \ J$. The frequency of oscillation 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: