A particle starts simple harmonic motion from the mean position. Its amplitude is a and total energy E . At one instant its kinetic energy is 3E/4 . Its displacement at that instant is

Options

A. `a//sqrt 2`
B. a/2
C. `a /sqrt(3//2)`
D. `a//sqrt3`

Answer

To solve the problem step by step, we will use the principles of simple harmonic motion (SHM) and the relationships between kinetic energy, potential energy, and total energy.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle is in simple harmonic motion (SHM) starting from the mean position.
- The amplitude of the motion is \( A \).
- The total energy of the system is \( E \).
- At a certain instant, the kinetic energy (KE) is given as \( \frac{3E}{4} \).
- We need to find the displacement \( x \) at that instant.

2. **Total Energy in SHM**:
- The total energy \( E \) in SHM is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
- Here, \( m \) is the mass of the particle, \( \omega \) is the angular frequency, and \( A \) is the amplitude.

3. **Kinetic Energy in SHM**:
- The kinetic energy \( KE \) at any point in SHM is given by:
\[ KE = \frac{1}{2} m v^2 \]
- The velocity \( v \) in SHM can be expressed as:
\[ v = A \omega \cos(\omega t) \]
- Therefore, the kinetic energy can be rewritten as:
\[ KE = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \]

4. **Setting Up the Equation**:
- We know that at a certain instant, the kinetic energy is \( \frac{3E}{4} \):
\[ \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = \frac{3}{4} E \]
- Substituting \( E \) into the equation:
\[ \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = \frac{3}{4} \left(\frac{1}{2} m A^2 \omega^2\right) \]

5. **Simplifying the Equation**:
- Canceling \( \frac{1}{2} m A^2 \omega^2 \) from both sides (assuming it is not zero):
\[ \cos^2(\omega t) = \frac{3}{4} \]
- Taking the square root:
\[ \cos(\omega t) = \frac{\sqrt{3}}{2} \]

6. **Finding the Angle**:
- The angle \( \omega t \) corresponding to \( \cos(\omega t) = \frac{\sqrt{3}}{2} \) is:
\[ \omega t = \frac{\pi}{6} \quad \text{(or any equivalent angle)} \]

7. **Calculating Displacement**:
- The displacement \( x \) at this instant can be calculated using:
\[ x = A \sin(\omega t) \]
- Substituting \( \omega t = \frac{\pi}{6} \):
\[ x = A \sin\left(\frac{\pi}{6}\right) = A \cdot \frac{1}{2} = \frac{A}{2} \]

8. **Final Answer**:
- Since the amplitude is given as \( a \), the displacement at that instant is:
\[ x = \frac{a}{2} \]

### Summary:
The displacement of the particle at the instant when its kinetic energy is \( \frac{3E}{4} \) is \( \frac{a}{2} \).