Two pendulums of length 1.21 m and 1.0 m starts vibrationg. At some instant, the two are in the mean position in same phase. After how many vibrations of the longer pendulum, the two will be in phase ?

Options

A. 10
B. 11
C. 20
D. 21

Answer

To solve the problem, we need to find out after how many vibrations of the longer pendulum (length 1.21 m) the two pendulums will be in phase again, given that they start in phase.

### Step-by-step Solution:

1. **Identify the lengths of the pendulums:**
- Length of the first pendulum (longer): \( L_1 = 1.21 \, \text{m} \)
- Length of the second pendulum (shorter): \( L_2 = 1.0 \, \text{m} \)

2. **Use the formula for the time period of a simple pendulum:**
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( T \) is the time period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.

3. **Calculate the time periods of both pendulums:**
- For the first pendulum:
\[ T_1 = 2\pi \sqrt{\frac{1.21}{g}} \]
- For the second pendulum:
\[ T_2 = 2\pi \sqrt{\frac{1.0}{g}} \]

4. **Find the ratio of the time periods:**
\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{1.21/g}}{2\pi \sqrt{1.0/g}} = \sqrt{\frac{1.21}{1.0}} = \sqrt{1.21} = \frac{11}{10} \]

5. **Express the relationship between the number of oscillations:**
Let \( n_1 \) be the number of oscillations of the longer pendulum and \( n_2 \) be the number of oscillations of the shorter pendulum. From the ratio of the time periods:
\[ n_1 T_1 = n_2 T_2 \]
Substituting the time periods:
\[ n_1 \cdot 2\pi \sqrt{\frac{1.21}{g}} = n_2 \cdot 2\pi \sqrt{\frac{1.0}{g}} \]
Simplifying gives:
\[ n_1 \sqrt{1.21} = n_2 \]
or
\[ n_2 = \frac{11}{10} n_1 \]

6. **Determine the smallest integers \( n_1 \) and \( n_2 \) such that both are integers:**
The smallest integers that satisfy the ratio \( n_2 = \frac{11}{10} n_1 \) would be:
- If \( n_1 = 10 \), then \( n_2 = 11 \).

7. **Conclusion:**
After 10 vibrations of the longer pendulum, the two pendulums will be in phase again.

### Final Answer:
The two pendulums will be in phase again after **10 vibrations** of the longer pendulum.

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