Question

For a particle P performing S.H.M., when displacement is 'x', potential energy and restoring force acting on it is denoted by 'E' and 'F' respectively. The relation between x, E and F is

• A. \(\frac{2\text{E}}{\text{F}}-x = 0\)
• B. \(\frac{2\text{E}}{\text{F}}+x = 0\)
• C. \(\frac{2\text{F}}{\text{x}}+\text{F} = 0\)
• D. \(\frac{2\text{E}}{\text{x}}-\text{F} = 0\)

Hint:

Use E = (1/2) k x^2 and F = -k x.

Answer 1

The Correct Option is B

Step 1: Pick a point:
At displacement $x$, $E = \tfrac12kx^2$ and $F = -kx$.

Step 2: Eliminate k:
$k = -F/x$. So $E = \tfrac12\left(-\dfrac Fx\right)x^2 = -\dfrac{Fx}{2}$.

Step 3: Rearrange:
$2E = -Fx$, so $\dfrac{2E}{F} + x = 0$, option (B).

Final Answer:
The relation is 2E/F + x = 0. \[ \boxed{\text{(B) }\dfrac{2E}{F}+x=0} \]