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: Understanding the Concept:
In SHM the restoring force is \(F = -kx\) and the potential energy is \(E = \tfrac12kx^2\).

Step 2: Divide the two:
\[ \frac EF = \frac{\tfrac12kx^2}{-kx} = -\frac x2 \]

Step 3: Rearrange:
\[ \frac{2E}{F} = -x \Rightarrow \frac{2E}{F} + x = 0 \]

Step 4: Check:
Option (B). Option (A) has the wrong sign. Options (C) and (D) are not consistent: in (D), \(\tfrac{2E}{x} = kx\) while \(F = -kx\), so \(\tfrac{2E}{x} - F\) would be \(2kx\), not zero.

Final Answer:
Dividing E by F gives 2E/F = -x. \[ \boxed{\text{(B) }\dfrac{2E}{F}+x=0} \]