Question

A body is executing S.H.M. under the action of force having maximum magnitude $50\text{ N}$. When its energy is half kinetic and half potential, the magnitude of the force acting on the particle is

• A. $\frac{25}{\sqrt{2}}\text{ N}$
• B. $50\text{ N}$
• C. $25\text{ N}$
• D. $25\sqrt{2}\text{ N}$

Hint:

Since force scales linearly with displacement ($F \propto x$) and potential energy scales quadratically with displacement ($P.E. \propto x^2$), halving the energy requires a displacement change of exactly $\frac{1}{\sqrt{2}}$. Therefore, the force must also drop by a factor of $\frac{1}{\sqrt{2}}$. Recognizing that $\frac{50}{\sqrt{2}} = 25\sqrt{2}$ yields the answer immediately.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
An object undergoes Simple Harmonic Motion (S.H.M.) under a restoring force whose peak magnitude at the maximum displacement limits is $F_{max} = 50\text{ N}$.
We need to determine the instantaneous magnitude of the restoring force $F'$ acting on the particle at the exact point along its path where its mechanical energy is split equally between kinetic energy ($K.E.$) and potential energy ($P.E.$).

Step 2: Key Formula or Approach:
1. The restoring force in S.H.M. is directly proportional to displacement: $F = kx$. The maximum force occurs at full amplitude ($x = A$), so $F_{max} = kA$.
2. The potential energy at any displacement $x$ is $P.E. = \frac{1}{2}kx^2$, and the total mechanical energy of the system is $E_{total} = \frac{1}{2}kA^2$.
3. When energy is shared half-and-half, the potential energy equals exactly half of the total energy: $P.E. = \frac{1}{2}E_{total}$.

Step 3: Detailed Explanation:
Let's set up the energy equation to determine the displacement $x$ for this specific state:
$$\frac{1}{2}kx^2 = \frac{1}{2}\left[\frac{1}{2}kA^2\right]$$ Cancel out the common multiplier $\frac{1}{2}k$ from both sides of the equation:
$$x^2 = \frac{A^2}{2} \implies x = \frac{A}{\sqrt{2}}$$ Now, calculate the restoring force $F'$ at this displacement value using our force relationship:
$$F' = kx = k\left(\frac{A}{\sqrt{2}}\right) = \frac{kA}{\sqrt{2}}$$ Substitute the maximum force identity ($F_{max} = kA = 50\text{ N}$) back into this expression:
$$F' = \frac{50}{\sqrt{2}}$$ Rationalize the denominator by multiplying the top and bottom by $\sqrt{2}$:
$$F' = \frac{50\sqrt{2}}{2} = 25\sqrt{2}\text{ N}$$

Step 4: Final Answer:
The magnitude of the force acting on the particle is $25\sqrt{2}\text{ N}$, which matches option (D).