Question

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

• A. 25 J
• B. 30 J
• C. 35 J
• D. 40 J

Answer 1

The correct option is D 40 J

$T=2\pi \sqrt{\frac{m}{K}}$
$T^2=4{\pi }^2\frac{m}{K}\Rightarrow K=\frac{4{\pi }^{2}m}{T^2}=\frac{4{\pi }^2\left(2\right)}{16}=\frac{{\pi }^2}{2}$
$\therefore$ In equilibrium, $U=\frac{1}{2}K{x}^2$
Where Kx = mg
$x=\frac{mg}{K}$
$\therefore U=\frac{1}{2}K\left(\frac{m^2{g}^2}{k^2}\right)=\frac{(mg{)}^2}{2k}={\left(\frac{mg}{\pi }\right)}^2=40J$