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: Force and energy in SHM.
The restoring force is $F = kx$, peaking at the amplitude where $F_{\max} = kA = 50\,\text{N}$. The total energy is $E = \tfrac{1}{2}kA^2$ and the potential energy at displacement $x$ is $\tfrac{1}{2}kx^2$.
Step 2: State the energy split.
Half kinetic and half potential means the potential energy equals half the total: \[ \tfrac{1}{2}kx^2 = \tfrac{1}{2}\left(\tfrac{1}{2}kA^2\right). \]
Step 3: Solve for the displacement.
Cancel $\tfrac{1}{2}k$: \[ x^2 = \frac{A^2}{2} \;\Rightarrow\; x = \frac{A}{\sqrt{2}}. \]
Step 4: Write the force at this point.
\[ F' = kx = \frac{kA}{\sqrt{2}}. \]
Step 5: Use the maximum force.
Since $kA = 50\,\text{N}$, \[ F' = \frac{50}{\sqrt{2}}. \]
Step 6: Rationalise.
\[ F' = \frac{50\sqrt{2}}{2} = 25\sqrt{2}\,\text{N}. \] That is option (D). \[ \boxed{F' = 25\sqrt{2}\,\text{N}} \]