Q108 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
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$\dfrac{2\text{E}}{\text{F}} - x = 0$
- Correct answerB$\dfrac{2\text{E}}{\text{F}} + x = 0$
- C$\dfrac{2\text{F}}{\text{x}} + \text{F} = 0$
- D$\dfrac{2\text{E}}{\text{x}} - \text{F} = 0$
Solution
For a particle performing Simple Harmonic Motion (S.H.M.), the restoring force $F$ acting on it at a displacement $x$ is given by:
$F = -kx$
The potential energy $E$ of the particle at displacement $x$ is given by:
$E = \dfrac{1}{2}kx^2$
Dividing the expression for potential energy by the expression for restoring force, we get:
$\dfrac{E}{F} = \dfrac{\dfrac{1}{2}kx^2}{-kx}$
$\dfrac{E}{F} = -\dfrac{x}{2}$
Rearranging the terms, we obtain:
$\dfrac{2E}{F} = -x$
$\dfrac{2E}{F} + x = 0$
Answer: $\dfrac{2\text{E}}{\text{F}} + x = 0$
$F = -kx$
The potential energy $E$ of the particle at displacement $x$ is given by:
$E = \dfrac{1}{2}kx^2$
Dividing the expression for potential energy by the expression for restoring force, we get:
$\dfrac{E}{F} = \dfrac{\dfrac{1}{2}kx^2}{-kx}$
$\dfrac{E}{F} = -\dfrac{x}{2}$
Rearranging the terms, we obtain:
$\dfrac{2E}{F} = -x$
$\dfrac{2E}{F} + x = 0$
Answer: $\dfrac{2\text{E}}{\text{F}} + x = 0$
Hence, the correct answer is option: 'B'.