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. `sqrt(3)(a)/(2)`
D. zero

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A particle starts SHM from the mean position (x = 0).
- The amplitude of the motion is given as \( a \).
- The total energy \( E \) of the system is given.
- At a certain instant, the kinetic energy \( K \) is \( \frac{3E}{4} \).
- 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 \( K \) at a displacement \( x \) is given by:
\[ K = \frac{1}{2} m \omega^2 a^2 - \frac{1}{2} m \omega^2 x^2 \]
- This can be rewritten as:
\[ K = \frac{1}{2} m \omega^2 (a^2 - x^2) \]

4. **Setting Up the Equation**:
- Given that \( K = \frac{3E}{4} \), we can substitute for \( E \):
\[ \frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{3}{4} \left( \frac{1}{2} m \omega^2 a^2 \right) \]
- Simplifying this, we get:
\[ a^2 - x^2 = \frac{3}{4} a^2 \]

5. **Solving for Displacement**:
- Rearranging the equation:
\[ x^2 = a^2 - \frac{3}{4} a^2 \]
\[ x^2 = \frac{1}{4} a^2 \]
- Taking the square root of both sides:
\[ x = \pm \frac{a}{2} \]

6. **Conclusion**:
- The displacement \( x \) at that instant is \( \frac{a}{2} \) or \( -\frac{a}{2} \).
- Since the options provided only include \( \frac{a}{2} \), we select that as the answer.

### Final Answer:
The displacement at that instant is \( \frac{a}{2} \).

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