Question

In the figure shown, the spring deflects by $\delta$ 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 of the spring mass system from position A to position B is

• A. $\frac{1}{2}k{x}^2$
• B. $\frac{1}{2}k{x}^2-mgx$
• C. $\frac{1}{2}k(x+\delta )$
• D. $\frac{1}{2}k{x}^2+mgx$

Answer 1

The correct option is A $\frac{1}{2}k{x}^2$

$\Delta \left(PE\right)=(PE{)}_B-(PE{)}_A$
$=\frac{1}{2}k(x+\delta {)}^2+0-[\frac{1}{2}k{\delta }^2+mgx]$

Taking reference datum at position B.
At equilibrium position i.e., at A.

$mg=k\delta$

$\Delta \left(PE\right)=\frac{1}{2}k{x}^2+\frac{1}{2}k{\delta }^2+xk\delta -\frac{1}{2}k{\delta }^2-mgx$

$\text{as}mg=k\delta$

$\Delta \left(PE\right)=\frac{1}{2}k{x}^2+\frac{1}{2}k{\delta }^2+mgx-\frac{1}{2}k{\delta }^2-mgx$

$=\frac{1}{2}k{x}^2$