A particle performs SHM about `x=0` such that at `t=0` it is at `x=0` and moving towards positive extreme. The time taken by it to go from `x=0` to `x=(A)/(2)` is____ time the time taken to go from `x=(A)/(2)` to `A`. The most suitable option for the blank space is

Options

A. 2
B. `(1)/(2)`
C. `(3A)/(T)`
D. `(A)/(2T)`

Explanation

To solve the problem, we need to analyze the motion of a particle performing Simple Harmonic Motion (SHM) and determine the time ratios for the specified intervals.

### Step-by-Step Solution:

1. **Understanding the SHM Setup**:
- The particle is performing SHM about the mean position \( x = 0 \).
- At \( t = 0 \), the particle is at \( x = 0 \) and moving towards the positive extreme position \( x = A \).

2. **Identifying Key Points**:
- Let’s denote:
- \( O \) as the position \( x = 0 \)
- \( P \) as the position \( x = \frac{A}{2} \)
- \( Q \) as the position \( x = A \)

3. **Time to Move from \( O \) to \( P \)**:
- The time taken to go from \( O \) to \( P \) (from \( x = 0 \) to \( x = \frac{A}{2} \)) can be calculated using the relationship between the angle in circular motion and the sine function.
- The sine of the angle \( \theta_1 \) corresponding to \( x = \frac{A}{2} \) is given by:
\[ \sin(\theta_1) = \frac{x}{A} = \frac{1}{2} \]
- Therefore, \( \theta_1 = \frac{\pi}{6} \).

4. **Calculating Time from \( O \) to \( P \)**:
- The time taken to cover the angle \( \theta_1 \) is:
\[ t_{OP} = \frac{\theta_1}{\omega} = \frac{\frac{\pi}{6}}{\omega} \]
- Here, \( \omega = \frac{2\pi}{T} \) (where \( T \) is the time period of the SHM).
- Thus, substituting \( \omega \):
\[ t_{OP} = \frac{\frac{\pi}{6}}{\frac{2\pi}{T}} = \frac{T}{12} \]

5. **Time to Move from \( P \) to \( Q \)**:
- The total time to go from \( O \) to \( Q \) (from \( x = 0 \) to \( x = A \)) is \( \frac{T}{4} \).
- Therefore, the time taken to go from \( P \) to \( Q \) is:
\[ t_{PQ} = t_{OQ} - t_{OP} = \frac{T}{4} - \frac{T}{12} \]
- To perform the subtraction, convert \( \frac{T}{4} \) to a fraction with a denominator of 12:
\[ t_{PQ} = \frac{3T}{12} - \frac{T}{12} = \frac{2T}{12} = \frac{T}{6} \]

6. **Finding the Ratio of Times**:
- Now, we can find the ratio of the time taken from \( O \) to \( P \) and from \( P \) to \( Q \):
\[ \frac{t_{OP}}{t_{PQ}} = \frac{\frac{T}{12}}{\frac{T}{6}} = \frac{1/12}{1/6} = \frac{1}{2} \]

### Conclusion:
The time taken by the particle to go from \( x = 0 \) to \( x = \frac{A}{2} \) is \( \frac{1}{2} \) times the time taken to go from \( x = \frac{A}{2} \) to \( x = A \).

### Final Answer:
The most suitable option for the blank space is **1/2**.

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