Two simple pendulum of lengths 1 m and 9 m respectively are both given a displacement in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed number of oscillations are

Options

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

Explanation

To solve the problem of when the two simple pendulums will be in phase after being displaced, we can follow these steps:

### Step 1: Understand the relationship between the length of the pendulum and its time period.
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.

### Step 2: Calculate the time periods for both pendulums.
For the first pendulum (length \( L_1 = 1 \, \text{m} \)):
\[ T_1 = 2\pi \sqrt{\frac{1}{g}} \]

For the second pendulum (length \( L_2 = 9 \, \text{m} \)):
\[ T_2 = 2\pi \sqrt{\frac{9}{g}} = 2\pi \sqrt{9} \cdot \frac{1}{\sqrt{g}} = 6\pi \sqrt{\frac{1}{g}} = 3T_1 \]

### Step 3: Determine the ratio of the time periods.
From the calculations, we find that:
\[ T_2 = 3T_1 \]
This means the second pendulum takes three times longer to complete one oscillation compared to the first pendulum.

### Step 4: Find the number of oscillations completed by the shorter pendulum when they are in phase.
To find when they will be in phase, we need to find a common time where both pendulums have completed an integer number of oscillations. Let \( n_1 \) be the number of oscillations of the first pendulum and \( n_2 \) be the number of oscillations of the second pendulum.

The time taken for \( n_1 \) oscillations of the first pendulum is:
\[ t = n_1 T_1 \]
The time taken for \( n_2 \) oscillations of the second pendulum is:
\[ t = n_2 T_2 = n_2 (3T_1) \]

Setting these equal gives:
\[ n_1 T_1 = n_2 (3T_1) \]
Dividing both sides by \( T_1 \) (assuming \( T_1 \neq 0 \)):
\[ n_1 = 3n_2 \]

### Step 5: Determine the smallest integers \( n_1 \) and \( n_2 \).
The smallest integers that satisfy this equation are \( n_1 = 3 \) and \( n_2 = 1 \). This means that after the shorter pendulum (1 m) completes 3 oscillations, the longer pendulum (9 m) will have completed 1 oscillation.

### Conclusion:
Thus, the shorter pendulum will complete **3 oscillations** before both pendulums are again in phase.

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