Question

A body of mass $2kg$ suspended through a vertical spring executes simple harmonic motion of period $4s$. When the oscillations are paused and the body is kept in equilibrium, what will be the potential energy stored in the spring?

Answer 1

The time period of a spring can be found by taking the square root of the ratio of the mass of the body to the spring constant. Find out the value of spring constant from this equation. The force will be equivalent to the product of the spring constant and the displacement of the body. Using all these relations find out the potential energy of the spring.

Complete step by step answer:
It has been given that the mass of the body suspended through a vertical spring can be given as,
$m=2kg$
Time period of the simple harmonic motion can be written as,
$T=4s$
We know the time period $T$ of mass $m$ on a spring of spring constant $k$can be found by taking the square root of the ratio of the mass of the body to the spring constant. This can be written as,
$T=2\pi \sqrt{\dfrac{m}{k}}$
Rearranging this equation in terms of the spring constant can be shown as,
$k=\dfrac{4{{\pi }^{2}}m}{{{T}^{2}}}$
Substituting the value of time period in it will give,
$k=\dfrac{4{{\pi }^{2}}\times 2}{16}=4.93N$
Also the force will be equivalent to the product of the mass of the body to the acceleration due to gravity. Otherwise the force can be shown as,
$F=kx$…….. (1)
Where $x$ be the displacement.
Also the potential energy of the spring can be found as,
$U=\dfrac{1}{2}k{{x}^{2}}$
Now from the equation (1) we can write that,
$x=\dfrac{2\times 10}{4.93}=4.05m$
Thus the potential energy stored in the spring can be found by the substitution of these values in the equation. That is,
$U=\dfrac{1}{2}\left( 4.93\times {{4.05}^{2}} \right)=40.43J$
Hence the potential energy of the spring has been obtained.

Note: Elastic potential energy is defined as the potential energy stored because of the deformation of an elastic body like the stretching of a spring. It will be equivalent to the work done to stretch the spring, which will be dependent on the spring constant as well as the stretched distance.