Question

A body of mass 2 kg suspended through a vertical spring executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring.

Answer 1

It is given that:
Mass of the body, m = 2 kg
Time period of the spring mass system, T = 4 s
The time period for spring-mass system is given as,
$T=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$
where k is the spring constant.

On substituting the respective values, we get:
$$\begin{gathered} \Rightarrow 4=2\mathrm{\pi }\sqrt{\frac{2}{k}} \\ \Rightarrow 2=\mathrm{\pi }\sqrt{\frac{2}{k}} \\ \Rightarrow 4={\mathrm{\pi }}^2\left(\frac{2}{k}\right) \\ \Rightarrow k=\frac{2{\mathrm{\pi }}^2}{4} \\ =\frac{{\mathrm{\pi }}^2}{2}=5\mathrm{N}/\mathrm{m} \end{gathered}$$

As the restoring force is balanced by the weight, we can write:
F = mg = kx
$\Rightarrow x=\frac{mg}{k}=\frac{2\times 10}{5}=4$

∴ Potential Energy $\left(U\right)$ of the spring is,
$$\begin{gathered} U=\left(\frac{1}{2}\right)k{x}^2=\frac{1}{2}\times 5\times 16 \\ =5\times 8=40\mathrm{J} \end{gathered}$$