Two simple pendulum of length `1m` and `16m` respectively are both given small displacement in the same direction of the same instant. They will be phase after one shorter pendulum has complated a oscillations. The value of `n` is

Options

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

Explanation

To solve the problem, we need to determine how many oscillations the shorter pendulum (length 1m) completes before the two pendulums are in phase again.

### Step-by-Step Solution:

1. **Identify the Lengths of the Pendulums:**
- Let the length of the first pendulum (shorter) be \( l_1 = 1 \, \text{m} \).
- Let the length of the second pendulum (longer) be \( l_2 = 16 \, \text{m} \).

2. **Calculate the Time Periods of the Pendulums:**
- The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{l}{g}} \]
- For the first pendulum:
\[ T_1 = 2\pi \sqrt{\frac{1}{g}} \]
- For the second pendulum:
\[ T_2 = 2\pi \sqrt{\frac{16}{g}} = 2\pi \cdot 4 \sqrt{\frac{1}{g}} = 4T_1 \]

3. **Determine the Relationship Between the Time Periods:**
- From the above calculations, we find:
\[ T_2 = 4T_1 \]

4. **Calculate the Number of Oscillations:**
- Let \( n \) be the number of oscillations completed by the shorter pendulum when both pendulums are in phase again.
- The time taken for \( n \) oscillations of the shorter pendulum is:
\[ t = nT_1 \]
- The time taken for the longer pendulum to complete its oscillations is:
\[ t = \frac{t}{T_2} = \frac{nT_1}{4T_1} = \frac{n}{4} \]

5. **Set the Condition for Being in Phase:**
- For the two pendulums to be in phase, the time taken for the longer pendulum must be an integer multiple of its period:
\[ t = mT_2 \quad \text{for some integer } m \]
- Since \( T_2 = 4T_1 \), we can write:
\[ nT_1 = m(4T_1) \]
- Dividing both sides by \( T_1 \):
\[ n = 4m \]

6. **Determine the Value of \( n \):**
- The smallest integer value for \( m \) is 1, which gives:
\[ n = 4 \times 1 = 4 \]

### Conclusion:
The value of \( n \) is \( 4 \). Therefore, the shorter pendulum will complete 4 oscillations before both pendulums are in phase again.