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

$m=2kg,T=4sec$

$T=2\pi \sqrt{\left(\frac{m}{k}\right)}$

$\Rightarrow 4=2\pi \sqrt{\frac{2}{k}}$

$\Rightarrow 2=\pi \sqrt{\frac{2}{k}}$

$\Rightarrow 4={\pi }^2\left(\frac{2}{k}\right)$

$\Rightarrow k=\frac{2{\pi }^2}{4}$

$=\frac{{\pi }^2}{2}=5N/m$

But, we know that

$F=mg=kx$

$\Rightarrow x=\frac{mg}{k}=\frac{2\times 10}{5}=4$

$\therefore$ Potential Energy $=\left(\frac{1}{2}\right)k{x}^2=\frac{1}{2}\times 5\times 16$

$=5\times 8=40j$