- In the figure shown, the spring deflects by d to position A (the equilibrium position) when a mass m is kept on it. During free vibration, the mass is at position B at some instant. The change in potential energy o the spring mass system from position A to position B is
(a) 
$\frac{1}{2}kx^2$ (b)$\frac{1}{2}kx^2-mgx$ 
(c) 
$\frac{1}{2}k\left (x+\delta \right )^{2}$ (d) $\frac{1}{2}kx^2+mgx$