The potential energy of a particle of mass $100 \,g$ moving along the $x$-axis is given by $U = 5x(x - 4)$,where $x$ is in metres. The period of oscillation is .................

$A$ body executes $SHM$ under the action of force $F_1$ with time period $T_1$. If the force is changed to $F_2$,it executes $SHM$ with time period $T_2$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,its time period is:

$A$ body of mass $4.9 \, kg$ hangs from a spring and oscillates with a period $0.5 \, s$. On the removal of the body, the spring is shortened by (take $g=10 \, m/s^2, \pi^2=10$)

$A$ horizontal spring executes $S.H.M.$ with amplitude $A_{1}$,when mass $m_{1}$ is attached to it. When it passes through the mean position,another mass $m_{2}$ is placed on it. Both masses move together with amplitude $A_{2}$. Therefore,the ratio $A_{2}: A_{1}$ is:

$A$ mass $0.9 \, kg$,attached to a horizontal spring,executes $SHM$ with an amplitude $A_{1}$. When this mass passes through its mean position,a smaller mass of $124 \, g$ is placed over it and both masses move together with amplitude $A_{2}$. If the ratio $\frac{A_{1}}{A_{2}}$ is $\frac{\alpha}{\alpha-1}$,then the value of $\alpha$ will be $......$

$A$ mass attached to a spring is free to oscillate,with angular velocity $\omega$,in a horizontal plane without friction or damping. It is pulled to a distance $x_{0}$ and pushed towards the centre with a velocity $v_{0}$ at time $t=0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_{0}$ and $v_{0}$. [Hint: Start with the equation $x=A \cos (\omega t+\theta)$ and note that the initial velocity is negative.]