Question

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

• A. $2EF - x^2 = 0$
• B. $2EF + x^2 = 0$
• C. $2EF + x = 0$
• D. $2EF - x = 0$

Hint:

You can verify this quickly using signs alone. Restoring force $F$ always acts in the opposite direction of displacement $x$, making their product $F \cdot x$ negative. Since potential energy $E$ is strictly positive, the ratio $\frac{2E}{F}$ must be negative. Adding $x$ (which balances the sign mismatch) is the only way to sum the terms to zero.

Answer 1

The Correct Option is C

Step 1: Identify the quantities.
For a particle in S.H.M. at displacement $x$, the potential energy is $E$ and the restoring force is $F$. We must connect $x$, $E$ and $F$.
Step 2: Write the restoring force.
The restoring force is proportional to displacement: $F = -kx$, where $k$ is the force constant.
Step 3: Write the potential energy.
The energy stored is $E = \frac{1}{2}kx^2$.
Step 4: Eliminate the constant $k$.
From the force equation, $k = -\dfrac{F}{x}$.
Step 5: Substitute into the energy equation.
$E = \frac{1}{2}\left(-\dfrac{F}{x}\right)x^2 = -\frac{1}{2}Fx$.
Step 6: Rearrange into option form.
Multiply by $2$: $2E = -Fx$. Dividing by $F$ gives $\dfrac{2E}{F} = -x$, i.e. $\dfrac{2E}{F} + x = 0$, written in the options as $2E/F + x = 0$, which is option (3).
\[ \boxed{\frac{2E}{F} + x = 0} \]