A paricle performs `S.H.M.` with time period `T`. The time taken by the particle to move from hall the amplitude to the maximum dispalcement is `T/2`

Options

A. `T/2`
B. `T/4`
C. `T/6`
D. `T/8`

Explanation

To solve the problem step by step, we need to analyze the motion of a particle performing Simple Harmonic Motion (SHM) and find the time taken to move from half the amplitude to the maximum displacement.

### Step-by-Step Solution:

1. **Understanding SHM**:
- In SHM, the displacement \( y \) of a particle can be described by the equation:
\[ y = A \sin(\omega t) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( t \) is the time.

2. **Identify Maximum Displacement**:
- The maximum displacement (amplitude) is \( A \).
- Half the amplitude is \( \frac{A}{2} \).

3. **Finding Time for Maximum Displacement**:
- At maximum displacement \( A \):
\[ A = A \sin(\omega t_1) \implies \sin(\omega t_1) = 1 \implies \omega t_1 = \frac{\pi}{2} \]
- Therefore, the time \( t_1 \) to reach maximum displacement is:
\[ t_1 = \frac{\pi}{2\omega} \]

4. **Finding Time for Half Amplitude**:
- For half the amplitude \( \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t_2) \implies \sin(\omega t_2) = \frac{1}{2} \]
- The angle where \( \sin(\theta) = \frac{1}{2} \) is \( \frac{\pi}{6} \):
\[ \omega t_2 = \frac{\pi}{6} \implies t_2 = \frac{\pi}{6\omega} \]

5. **Calculating Time Taken from Half Amplitude to Maximum Displacement**:
- The time taken to move from \( \frac{A}{2} \) to \( A \) is:
\[ \Delta t = t_1 - t_2 = \left(\frac{\pi}{2\omega}\right) - \left(\frac{\pi}{6\omega}\right) \]
- Finding a common denominator (which is \( 6\omega \)):
\[ \Delta t = \left(\frac{3\pi}{6\omega}\right) - \left(\frac{\pi}{6\omega}\right) = \frac{2\pi}{6\omega} = \frac{\pi}{3\omega} \]

6. **Relating Time to Time Period**:
- The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]
- Therefore, we can express \( \omega \) in terms of \( T \):
\[ \omega = \frac{2\pi}{T} \]
- Substituting \( \omega \) back into \( \Delta t \):
\[ \Delta t = \frac{\pi}{3} \cdot \frac{T}{2\pi} = \frac{T}{6} \]

### Final Answer:
The time taken by the particle to move from half the amplitude to the maximum displacement is:
\[ \Delta t = \frac{T}{6} \]