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

Step 1: Understanding the Concept:
Total energy ($TE$) in SHM is the sum of kinetic energy ($KE$) and potential energy ($PE$). Total energy is constant throughout the motion.
Step 2: Key Formula or Approach:
1. Total Energy: $TE = \frac{1}{2} k A^2$
2. Kinetic Energy: $KE = \frac{1}{2} k (A^2 - x^2)$
Step 3: Detailed Explanation:
Given $x = \frac{\sqrt{3}}{2} A$.
1. Find Kinetic Energy:
\[ KE = \frac{1}{2} k \left[ A^2 - \left( \frac{\sqrt{3}}{2} A \right)^2 \right] = \frac{1}{2} k \left( A^2 - \frac{3}{4} A^2 \right) = \frac{1}{2} k \left( \frac{1}{4} A^2 \right) \]
2. Relate to Total Energy:
\[ KE = \frac{1}{4} \left( \frac{1}{2} k A^2 \right) = \frac{1}{4} TE \]
\[ TE = 4 \times KE \]
Comparing with $TE = n \times KE$, we find $n = 4$.
Step 4: Final Answer:
The value of $n$ is 4.