Question

For a particle performing S.H.M.; the total energy is ' $n$ ' times the kinetic energy, when the displacement of a particle from mean position is $\frac{\sqrt{3}{2}A$, where A is the amplitude of S.H.M. The value of ' $n$ ' is

• A. 2
• B. 3
• C. 4
• D. 6

Hint:

In SHM: \[ K=\frac12 k(A^2-x^2) \] At any position, total energy remains constant and equals \(\frac12 kA^2\).

Answer 1

The Correct Option is C

Concept:
In SHM: \[ E=\frac12 kA^2 \] and potential energy at displacement \(x\) is: \[ U=\frac12 kx^2 \] So kinetic energy is: \[ K=E-U=\frac12 k(A^2-x^2) \] ip

Step 1: Substitute the given displacement.
Given: \[ x=\frac{\sqrt{3}}{2}A \] So, \[ x^2=\frac34 A^2 \] ip

Step 2: Find kinetic energy at that position.
\[ K=\frac12 k\left(A^2-\frac34 A^2\right) \] \[ K=\frac12 k\left(\frac14 A^2\right) =\frac18 kA^2 \] ip

Step 3: Find the ratio \(E/K\).
\[ E=\frac12 kA^2 \] Therefore, \[ \frac{E}{K} = \frac{\frac12 kA^2}{\frac18 kA^2} =4 \] So, \[ E=4K \] Hence, \[ n=4 \] ip Hence, the correct answer is:
\[ \boxed{(C)\ 4} \]