When a mass $m$ is attached to a spring,it extends by $0.2 \ m$. The mass $m$ is given a slight additional extension and released. Find the time period (take $g = \pi^2 \ m/s^2$).

How does the period of oscillation depend on the mass of the block attached to the end of a spring?

When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released,its motion is described by $y(t) = y_{0} \sin^{2} \omega t$,where $y$ is measured from the lower end of the unstretched spring. Then $\omega$ is

Springs of spring constants $K, 2K, 4K, 8K, \dots$ are connected in series. $A$ mass of $40 \, g$ is attached to the lower end of the last spring and the system is allowed to vibrate. What is the time period of oscillation in seconds? (Given $K = 2 \, N/cm$)

$A$ sphere of mass $m$ is attached to the lower end of a light vertical spring with force constant $k$. The upper end of the spring is fixed. The sphere is released from rest at the natural (unstretched) length of the spring and comes to rest again after falling through a distance $x$.

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 .................