Question 359: Oscillations

Question

For a particle performing SHM when displacement is $$x$$, the potential energy and restoring force acting on it is denoted by $$E$$ and $$F$$, respectively. The relation between $$x, E$$ and $$F$$ is

Accepted Answer

For a particle performing Simple Harmonic Motion (SHM), the potential energy (PE) at a displacement $$x$$ from the equilibrium position is given by
$$E = \frac{1}{2} k x^2$$
where $$k$$ is the spring constant.
The restoring force $$F$$ acting on the particle is given by Hooke's law, which states that $$F = -kx$$. The negative sign indicates that the force is directed towards the equilibrium position, opposing the displacement.
To find the relation between $$x, E$$, and $$F$$, we can express $$k$$ from the equation of force:
$$F = -kx \Rightarrow k = -\frac{F}{x}$$
Substituting $$k$$ into the equation for potential energy:
$$E = \frac{1}{2} \left(-\frac{F}{x}\right) x^2 = \frac{1}{2} \cdot \frac{F}{x} \cdot x^2$$
We simplify this to get:
$$E = \frac{1}{2}Fx$$
To find a relational formula involving $$E$$, $$F$$, and $$x$$, we can solve the equation $$E = \frac{1}{2}Fx$$ for one variable in terms of the others. However, let's examine the given options instead. Rearranging the equation for $$E$$ to isolate terms similar to the options given, we obtain:
$$2E = Fx$$
Dividing both sides of this equation by $$F$$ gives us:
$$\frac{2E}{F} = x$$
Rearranging the terms to match the given options, we get:
$$\frac{2E}{F} - x = 0$$
Therefore, the correct relation between $$x, E$$, and $$F$$ is given by Option B:
$$\frac{2E}{F}-x=0$$

Answer

$$\frac{E}{F}+x=0$$

Answer

$$\frac{2 E}{F}-x=0$$

Answer

$$\frac{2 E}{F}+x=0$$

Answer

$$\frac{E}{F}-x=0$$

Oscillations

For a particle performing SHM when displacement is $x, the potential energy and restoring force acting on it is denoted by E and F, respectively. The relation between x, E and F$ is

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