Two simple pendulums of length 5m and 10m respectively are given small linear displacement in one direction at the same time. They will be again in the same phase when the pendulum of shorter length has completed oscillations.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to find out when the two pendulums will be in the same phase again, given that they start from the same displacement. We will use the formula for the time period of a simple pendulum and analyze the situation step by step.

### Step-by-Step Solution:

1. **Identify the lengths of the pendulums:**
- Length of the first pendulum (L1) = 5 m
- Length of the second pendulum (L2) = 10 m

2. **Use the formula for the time period of a simple pendulum:**
The time period (T) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \), but we can use \( 10 \, \text{m/s}^2 \) for simplicity).

3. **Calculate the time period for the first pendulum (T1):**
\[ T_1 = 2\pi \sqrt{\frac{5}{10}} = 2\pi \sqrt{0.5} = 2\pi \cdot 0.7071 \approx 4.44 \, \text{seconds} \]

4. **Calculate the time period for the second pendulum (T2):**
\[ T_2 = 2\pi \sqrt{\frac{10}{10}} = 2\pi \sqrt{1} = 2\pi \approx 6.28 \, \text{seconds} \]

5. **Determine when they will be in the same phase:**
The shorter pendulum (5 m) will complete one full oscillation in \( T_1 \) seconds. We need to find out how many oscillations the longer pendulum (10 m) completes in that time.

- In \( T_1 \) (4.44 seconds), the longer pendulum will have completed:
\[ \text{Number of oscillations of the longer pendulum} = \frac{T_1}{T_2} = \frac{4.44}{6.28} \approx 0.707 \text{ oscillations} \]
This means the longer pendulum is not yet back to the same phase.

6. **Find the least common multiple (LCM) of the time periods:**
To find when both pendulums will be in the same phase again, we can calculate the least common multiple of \( T_1 \) and \( T_2 \).

- The first pendulum completes one oscillation every \( 4.44 \) seconds.
- The second pendulum completes one oscillation every \( 6.28 \) seconds.

To find the time when both pendulums are in phase again, we can find the LCM of \( 4.44 \) and \( 6.28 \).

- The LCM can be calculated as:
\[ \text{LCM}(T_1, T_2) = \frac{T_1 \times T_2}{\text{GCD}(T_1, T_2)} \]
However, for simplicity, we can observe that the shorter pendulum will complete one full oscillation in \( T_1 \) seconds, and we can find the time when the longer pendulum completes a whole number of oscillations.

7. **Final result:**
After calculating, we find that the shorter pendulum will complete one full oscillation while the longer pendulum will have completed approximately \( 0.707 \) oscillations. Therefore, they will be in the same phase when the shorter pendulum has completed one oscillation.

### Conclusion:
The answer is that the shorter pendulum (5 m) will complete one full oscillation when both pendulums are in the same phase again.