The kinetic energy of SHM is 1/n time its potential energy. If the amplitude of the SHM be A, then what is the displacement of the particle ?

Options

A. nA
B. `sqrt((n+1)/(n))`
C. `sqrt((n)/(n+1))`
D. `(A)/(n)`

Explanation

To solve the problem, we need to find the displacement \( x \) of a particle in Simple Harmonic Motion (SHM) given that the kinetic energy (KE) is \( \frac{1}{n} \) times the potential energy (PE). The amplitude of SHM is given as \( A \).

### Step-by-Step Solution:

1. **Understand the Kinetic and Potential Energy in SHM**:
- The kinetic energy \( KE \) of a particle in SHM is given by:
\[ KE = \frac{1}{2} m v^2 \]
- The potential energy \( PE \) is given by:
\[ PE = \frac{1}{2} k x^2 \]
- Here, \( v \) is the velocity of the particle, \( m \) is the mass, \( k \) is the spring constant, and \( x \) is the displacement from the mean position.

2. **Relate Velocity to Displacement**:
- The velocity \( v \) in SHM can be expressed as:
\[ v = \sqrt{\omega^2 A^2 - \omega^2 x^2} = \omega \sqrt{A^2 - x^2} \]
- Where \( \omega \) is the angular frequency.

3. **Substituting Velocity into Kinetic Energy**:
- Substitute \( v \) into the kinetic energy formula:
\[ KE = \frac{1}{2} m (\omega \sqrt{A^2 - x^2})^2 = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

4. **Substituting Spring Constant**:
- The spring constant \( k \) can be expressed in terms of \( \omega \):
\[ k = m \omega^2 \]
- Thus, the potential energy becomes:
\[ PE = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2 \]

5. **Setting Up the Equation**:
- According to the problem, the kinetic energy is \( \frac{1}{n} \) times the potential energy:
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{n} \left( \frac{1}{2} m \omega^2 x^2 \right) \]
- Cancel \( \frac{1}{2} m \omega^2 \) from both sides:
\[ A^2 - x^2 = \frac{1}{n} x^2 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ A^2 = x^2 + \frac{1}{n} x^2 = \left(1 + \frac{1}{n}\right) x^2 \]
- This simplifies to:
\[ A^2 = \frac{n+1}{n} x^2 \]

7. **Solving for Displacement \( x \)**:
- Rearranging for \( x^2 \):
\[ x^2 = \frac{n}{n+1} A^2 \]
- Taking the square root gives:
\[ x = A \sqrt{\frac{n}{n+1}} \]

### Final Answer:
The displacement of the particle is:
\[ x = A \sqrt{\frac{n}{n+1}} \]