The time period of a particle executing `SHM` is `T`. After a time `T//6` after it passes its mean position, then at`t = 0` its :

Options

A. displacement will be one-half of its amplitude
B. velocity will be one-half of its maximum velocity
C. kinetic energy `= 1//3` (potential energy)
D. acceleration will be `(sqrt(3))/(2)` times of its maximum acceleration

Explanation

To solve the problem step by step, we will analyze the motion of a particle executing Simple Harmonic Motion (SHM) with a given time period \( T \). We will determine the displacement, velocity, acceleration, and kinetic energy of the particle at \( t = 0 \) after it has passed the mean position at \( t = \frac{T}{6} \).

### Step-by-Step Solution

1. **Understanding the Motion**:
- The time period of the particle is \( T \).
- The particle passes through the mean position at \( t = \frac{T}{6} \). At this point, the displacement \( x = 0 \).

2. **Finding the Phase Constant**:
- The general equation for displacement in SHM is given by:
\[ x(t) = A \sin(\omega t + \phi) \]
- Here, \( \omega = \frac{2\pi}{T} \).
- At \( t = \frac{T}{6} \), since \( x = 0 \):
\[ 0 = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{6} + \phi\right) \]
- This simplifies to:
\[ 0 = A \sin\left(\frac{\pi}{3} + \phi\right) \]
- For the sine function to be zero, the argument must be \( n\pi \) (where \( n \) is an integer). Thus:
\[ \frac{\pi}{3} + \phi = n\pi \]
- Solving for \( \phi \) gives:
\[ \phi = n\pi - \frac{\pi}{3} \]
- For \( n = 1 \), we get:
\[ \phi = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
- For \( n = 0 \), we get:
\[ \phi = -\frac{\pi}{3} \]
- We will use \( \phi = -\frac{\pi}{3} \).

3. **Calculating Displacement at \( t = 0 \)**:
- Now, we calculate the displacement at \( t = 0 \):
\[ x(0) = A \sin\left(\omega \cdot 0 + \phi\right) = A \sin\left(-\frac{\pi}{3}\right) \]
- Since \( \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \):
\[ x(0) = -A \frac{\sqrt{3}}{2} \]

4. **Calculating Velocity at \( t = 0 \)**:
- The velocity in SHM is given by:
\[ v(t) = A \omega \cos(\omega t + \phi) \]
- At \( t = 0 \):
\[ v(0) = A \omega \cos\left(0 + \phi\right) = A \omega \cos\left(-\frac{\pi}{3}\right) \]
- Since \( \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ v(0) = A \omega \cdot \frac{1}{2} = \frac{A \omega}{2} \]
- The maximum velocity \( V_{max} = A \omega \), hence:
\[ v(0) = \frac{V_{max}}{2} \]

5. **Calculating Acceleration at \( t = 0 \)**:
- The acceleration in SHM is given by:
\[ a(t) = -A \omega^2 \sin(\omega t + \phi) \]
- At \( t = 0 \):
\[ a(0) = -A \omega^2 \sin\left(-\frac{\pi}{3}\right) = A \omega^2 \frac{\sqrt{3}}{2} \]
- The maximum acceleration \( A_{max} = A \omega^2 \), hence:
\[ a(0) = \frac{A_{max} \sqrt{3}}{2} \]

6. **Calculating Kinetic Energy at \( t = 0 \)**:
- The kinetic energy is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Substituting \( v(0) = \frac{A \omega}{2} \):
\[ KE = \frac{1}{2} m \left(\frac{A \omega}{2}\right)^2 = \frac{1}{2} m \frac{A^2 \omega^2}{4} = \frac{m A^2 \omega^2}{8} \]

7. **Calculating Potential Energy at \( t = 0 \)**:
- The potential energy is given by:
\[ PE = \frac{1}{2} k x^2 \]
- Since \( k = m \omega^2 \) and \( x(0) = -A \frac{\sqrt{3}}{2} \):
\[ PE = \frac{1}{2} m \omega^2 \left(-A \frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} m \omega^2 \frac{3A^2}{4} = \frac{3m A^2 \omega^2}{8} \]

### Summary of Results
- Displacement at \( t = 0 \): \( x(0) = -A \frac{\sqrt{3}}{2} \)
- Velocity at \( t = 0 \): \( v(0) = \frac{V_{max}}{2} \)
- Acceleration at \( t = 0 \): \( a(0) = \frac{A_{max} \sqrt{3}}{2} \)
- Kinetic Energy at \( t = 0 \): \( KE = \frac{m A^2 \omega^2}{8} \)
- Potential Energy at \( t = 0 \): \( PE = \frac{3m A^2 \omega^2}{8} \)