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 of finding the time taken by a particle executing simple harmonic motion (SHM) with a time period \( T \) to complete \( \frac{5}{8} \) of an oscillation, we can follow these steps:

### Step 1: Understand the Motion
The particle starts from the mean position and executes SHM. The time period \( T \) is the time taken to complete one full oscillation.

### Step 2: Calculate the Time for \( \frac{5}{8} \) of an Oscillation
To find the time taken for \( \frac{5}{8} \) of the oscillation, we can use the relationship:
\[ \text{Time for } \frac{5}{8} \text{ of oscillation} = \frac{5}{8} \times T \]

### Step 3: Break Down the Oscillation
Since the particle starts from the mean position, we can analyze the motion in parts:
1. From the mean position to the right extreme (half the oscillation) takes \( \frac{T}{2} \).
2. From the right extreme back to the mean position takes another \( \frac{T}{2} \).
3. The particle then moves from the mean position to the left extreme (another half oscillation) takes another \( \frac{T}{2} \).
4. Finally, it returns from the left extreme to the mean position, completing the full oscillation.

### Step 4: Calculate the Time for \( \frac{5}{8} \) of the Oscillation
To find the time taken for \( \frac{5}{8} \) of the oscillation:
- The first \( \frac{1}{2} \) oscillation (from mean to right extreme and back) takes \( T \).
- The next \( \frac{1}{8} \) of the oscillation (from mean to \( \frac{A}{2} \)) needs to be calculated.

### Step 5: Calculate Time for \( \frac{1}{8} \) of the Oscillation
The time taken to move from the mean position to a displacement of \( \frac{A}{2} \) can be calculated using the formula:
\[ \text{Time} = \frac{T}{2\pi} \times \text{Angle in radians} \]
The angle corresponding to \( \frac{A}{2} \) can be found using the sine function:
\[ \sin(\phi) = \frac{A/2}{A} = \frac{1}{2} \implies \phi = \frac{\pi}{6} \]
Thus, the time taken to reach \( \frac{A}{2} \) is:
\[ t_{\frac{A}{2}} = \frac{\pi/6}{2\pi/T} = \frac{T}{12} \]

### Step 6: Total Time Calculation
Now, we can add the times:
\[ \text{Total time} = T + \frac{T}{12} = \frac{12T + T}{12} = \frac{13T}{12} \]

### Final Step: Conclusion
Thus, the total time taken by the particle to complete \( \frac{5}{8} \) of the oscillation is:
\[ \text{Total time} = \frac{7T}{12} \]