Two simple pendulums, having periods of `2s` and `3s` respectively, pass through the mean position simulaneously at a particular instant. They may be in phase after an interval of:

Options

A. `5s`
B. `6s`
C. `1s`
D. none of these above

Explanation

To solve the problem of when the two simple pendulums will be in phase again after passing through the mean position simultaneously, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Periods of the Pendulums**:
- Let the period of the first pendulum (T1) be 2 seconds.
- Let the period of the second pendulum (T2) be 3 seconds.

2. **Calculate the Angular Frequencies**:
- The angular frequency (ω) is related to the period (T) by the formula:
\[ \omega = \frac{2\pi}{T} \]
- For the first pendulum:
\[ \omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{2} = \pi \text{ rad/s} \]
- For the second pendulum:
\[ \omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{3} \text{ rad/s} \]

3. **Determine the Condition for Being in Phase**:
- The two pendulums will be in phase when the difference in their angular displacements is an integer multiple of \(2\pi\):
\[ (\omega_1 - \omega_2) \cdot t = 2n\pi \quad (n \text{ is an integer}) \]
- Rearranging gives:
\[ t = \frac{2n\pi}{\omega_1 - \omega_2} \]

4. **Calculate the Difference in Angular Frequencies**:
- Find \(\omega_1 - \omega_2\):
\[ \omega_1 - \omega_2 = \pi - \frac{2\pi}{3} = \frac{3\pi}{3} - \frac{2\pi}{3} = \frac{\pi}{3} \text{ rad/s} \]

5. **Substitute into the Time Equation**:
- Substitute \(\omega_1 - \omega_2\) into the time equation:
\[ t = \frac{2n\pi}{\frac{\pi}{3}} = 2n \cdot 3 = 6n \text{ seconds} \]

6. **Find the Smallest Positive Time**:
- For the smallest positive integer \(n = 1\):
\[ t = 6 \text{ seconds} \]

### Conclusion:
The two pendulums will be in phase again after an interval of **6 seconds**.