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.

### Step-by-Step Solution:

1. **Identify the lengths of the pendulums:**
- Length of the longer pendulum (L1) = 1.21 m
- Length of the shorter pendulum (L2) = 1.0 m

2. **Determine the formula for the time period of a 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.

3. **Calculate the time periods of both pendulums:**
- For the longer pendulum (T1):
\[ T_1 = 2\pi \sqrt{\frac{1.21}{g}} \]
- For the shorter pendulum (T2):
\[ T_2 = 2\pi \sqrt{\frac{1.0}{g}} \]

4. **Find the ratio of the time periods:**
\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{1.21}}{2\pi \sqrt{1.0}} = \sqrt{\frac{1.21}{1.0}} = \sqrt{1.21} = 1.1 \]
This means:
\[ T_1 = 1.1 T_2 \]

5. **Relate the number of vibrations:**
Let \(n_1\) be the number of vibrations of the longer pendulum and \(n_2\) be the number of vibrations of the shorter pendulum. The relationship between the vibrations can be expressed as:
\[ n_1 T_1 = n_2 T_2 \]
Substituting \(T_1\) in terms of \(T_2\):
\[ n_1 (1.1 T_2) = n_2 T_2 \]
Dividing both sides by \(T_2\):
\[ 1.1 n_1 = n_2 \]

6. **Express \(n_2\) in terms of \(n_1\):**
Rearranging gives:
\[ n_2 = 1.1 n_1 \]
This means for every 10 vibrations of the longer pendulum, the shorter pendulum will complete 11 vibrations.

7. **Find the least common multiple (LCM):**
To find when they will be in phase again, we need to find the smallest integers \(n_1\) and \(n_2\) that satisfy \(n_2 = 1.1 n_1\). The smallest integers that satisfy this are:
\[ n_1 = 10 \quad \text{and} \quad n_2 = 11 \]

8. **Conclusion:**
Therefore, after **10 vibrations** of the longer pendulum, the two pendulums will be in phase again.

### Final Answer:
After 10 vibrations of the longer pendulum, the two pendulums will be in phase.