Time period of a simple pendulum is T. The time taken to complete 5/8 oscillations starting from mean position is `(alpha)/(beta)T`. The value of `alpha` is …………. .

Answer

To solve the problem, we need to determine the time taken to complete \( \frac{5}{8} \) of an oscillation of a simple pendulum, given that the time period of the pendulum is \( T \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
The motion of a simple pendulum can be understood as simple harmonic motion (SHM). The total time period \( T \) is the time taken to complete one full oscillation (from the mean position to one extreme and back to the mean position, then to the other extreme and back).

2. **Dividing the Oscillation**:
We need to find the time taken for \( \frac{5}{8} \) of an oscillation. A full oscillation can be divided into 8 equal parts. Therefore, \( \frac{5}{8} \) of an oscillation corresponds to moving through 5 of these parts.

3. **Calculating the Time for Each Part**:
Since the total time for one complete oscillation is \( T \), the time for each part of the oscillation is:
\[ \text{Time for each part} = \frac{T}{8} \]

4. **Finding the Time for 5 Parts**:
The time taken to complete \( \frac{5}{8} \) of the oscillation is:
\[ \text{Time for } \frac{5}{8} \text{ oscillation} = 5 \times \frac{T}{8} = \frac{5T}{8} \]

5. **Comparing with Given Expression**:
We are given that the time taken to complete \( \frac{5}{8} \) oscillations is \( \frac{\alpha}{\beta} T \). From our calculation, we have:
\[ \frac{5T}{8} = \frac{\alpha}{\beta} T \]
By comparing both sides, we can see that:
\[ \frac{\alpha}{\beta} = \frac{5}{8} \]

6. **Identifying Values of \( \alpha \) and \( \beta \)**:
To find the value of \( \alpha \), we can let \( \beta = 1 \) (for simplicity), which gives us:
\[ \alpha = \frac{5}{8} \cdot 1 = 5 \]
However, since the problem asks for the value of \( \alpha \) in the context of the given expression, we can assume \( \beta = 1 \) and thus \( \alpha = 5 \).

### Final Answer:
The value of \( \alpha \) is \( 5 \).