The time period of a particle in simple harmonic motion is T. Assume potential energy at mean position to be zero. After a time of `(T)/(6)` it passes its mean position ,then at t=0 its,

Options

A. velocity will be one half its maximum velocity
B. displacement will be one half of its amplitude
C. Acceleration will be nearly `86^(@)` of its maximum acceleration
D. kinetic energy = potential energy

Answer

To solve the problem, we need to analyze the motion of a particle in simple harmonic motion (SHM) given that the time period is \( T \) and the potential energy at the mean position is zero. We are given that the particle passes through its mean position after a time of \( \frac{T}{6} \).

### Step-by-Step Solution:

1. **Understanding SHM**:
The displacement \( x \) of a particle in SHM can be represented as:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega = \frac{2\pi}{T} \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Mean Position**:
At the mean position, the displacement \( x \) is zero. We know that at \( t = \frac{T}{6} \), the particle passes through the mean position:
\[ x\left(\frac{T}{6}\right) = A \sin\left(\omega \frac{T}{6} + \phi\right) = 0 \]
This implies:
\[ \sin\left(\frac{2\pi}{T} \cdot \frac{T}{6} + \phi\right) = \sin\left(\frac{\pi}{3} + \phi\right) = 0 \]

3. **Finding Phase Constant**:
The sine function is zero when its argument is \( n\pi \) (where \( n \) is an integer). Therefore:
\[ \frac{\pi}{3} + \phi = n\pi \]
For the simplest case (taking \( n = 0 \)):
\[ \phi = -\frac{\pi}{3} \]

4. **Displacement Equation**:
Substituting \( \phi \) back into the displacement equation:
\[ x(t) = A \sin\left(\omega t - \frac{\pi}{3}\right) \]

5. **Finding Displacement at \( t = 0 \)**:
Now, we can find the displacement at \( t = 0 \):
\[ x(0) = A \sin\left(-\frac{\pi}{3}\right) = -A \frac{\sqrt{3}}{2} \]
This means the displacement at \( t = 0 \) is approximately \( -0.866A \).

6. **Finding Velocity**:
The velocity \( v(t) \) is given by the derivative of displacement:
\[ v(t) = \frac{dx}{dt} = A \omega \cos\left(\omega t - \frac{\pi}{3}\right) \]
At \( t = 0 \):
\[ v(0) = A \omega \cos\left(-\frac{\pi}{3}\right) = A \omega \cdot \frac{1}{2} \]
The maximum velocity \( V_{max} = A \omega \), thus:
\[ v(0) = \frac{1}{2} V_{max} \]

7. **Finding Acceleration**:
The acceleration \( a(t) \) is given by:
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \sin\left(\omega t - \frac{\pi}{3}\right) \]
At \( t = 0 \):
\[ a(0) = -A \omega^2 \sin\left(-\frac{\pi}{3}\right) = A \omega^2 \cdot \frac{\sqrt{3}}{2} \]
The maximum acceleration \( A_{max} = A \omega^2 \), thus:
\[ a(0) = \frac{\sqrt{3}}{2} A_{max} \approx 0.866 A_{max} \]

8. **Potential and Kinetic Energy**:
The potential energy \( U \) at \( x(0) \):
\[ U = \frac{1}{2} k x^2 = \frac{1}{2} k \left(-A \frac{\sqrt{3}}{2}\right)^2 = \frac{3}{8} k A^2 \]
The kinetic energy \( K \) at \( t = 0 \):
\[ K = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{1}{2} A \omega\right)^2 = \frac{1}{8} m A^2 \omega^2 = \frac{1}{8} k A^2 \]

### Conclusion:
- The displacement at \( t = 0 \) is approximately \( -0.866A \) (not half of the amplitude).
- The velocity at \( t = 0 \) is half of the maximum velocity.
- The acceleration at \( t = 0 \) is approximately \( 0.866 A_{max} \).
- The potential and kinetic energies are not equal.