A particle is executing SHM with time period T Starting from mean position, time taken by it to complete `(5)/(8)` oscillations is,

Options

A. `(T)/(12)`
B. `(T)/(6)`
C. `(5T)/(12)`
D. `(7T)/(12)`

Answer

To solve the problem, we need to determine the time taken by a particle executing Simple Harmonic Motion (SHM) to complete \( \frac{5}{8} \) of an oscillation, given that the time period of the oscillation is \( T \).

### Step-by-Step Solution:

1. **Understand the Time Period**: The time period \( T \) is the time taken to complete one full oscillation (from mean position to the maximum displacement and back to mean position).

2. **Calculate the Time for \( \frac{5}{8} \) Oscillations**:
- Since the time for one complete oscillation is \( T \), the time taken for \( \frac{5}{8} \) of an oscillation can be calculated as:
\[ \text{Time for } \frac{5}{8} \text{ oscillations} = \frac{5}{8} \times T \]

3. **Substituting the Values**:
- If we substitute the value of \( T \) into the equation, we get:
\[ \text{Time for } \frac{5}{8} \text{ oscillations} = \frac{5T}{8} \]

4. **Conclusion**:
- Therefore, the time taken by the particle to complete \( \frac{5}{8} \) of an oscillation is:
\[ \frac{5T}{8} \]

### Final Answer:
The time taken by the particle to complete \( \frac{5}{8} \) oscillations is \( \frac{5T}{8} \).

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