Two simple pendulums of lengths 1.44 m and 1 m start swinging together. After how many vibrations will they again start swinging together

Options

A. 5 oscillations of smaller pendulum
B. 6 oscillations of smaller pendulum
C. 4 oscillations of bigger pendulum
D. 6 oscillations of bigger pendulum

Answer

To solve the problem of when two simple pendulums of lengths 1.44 m and 1 m will swing together again, we can follow these steps:

### Step 1: Understand the Time Period Formula
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 \( 9.81 \, \text{m/s}^2 \)).

### Step 2: Calculate the Time Periods of Both Pendulums
For the first pendulum (length \( L_1 = 1.44 \, \text{m} \)):
\[ T_1 = 2\pi \sqrt{\frac{1.44}{g}} \]

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

### Step 3: Find the Ratio of the Time Periods
To find when they will swing together again, we can take the ratio of the time periods:
\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{1.44}{g}}}{2\pi \sqrt{\frac{1}{g}}} = \sqrt{\frac{1.44}{1}} = \sqrt{1.44} = 1.2 \]

This means that \( T_1 = 1.2 T_2 \).

### Step 4: Determine the 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 when they next align. Since they start together, we can express their relationship as:
\[ n_1 T_1 = n_2 T_2 \]

Substituting \( T_1 = 1.2 T_2 \):
\[ n_1 (1.2 T_2) = n_2 T_2 \]

Dividing both sides by \( T_2 \):
\[ 1.2 n_1 = n_2 \]

### Step 5: Find the Smallest Integer Values
To find the smallest integers \( n_1 \) and \( n_2 \) that satisfy this equation, we can set \( n_1 = 5 \) (the number of oscillations of the longer pendulum). Then:
\[ n_2 = 1.2 \times 5 = 6 \]

### Conclusion
Thus, the two pendulums will swing together again after:
- **5 oscillations of the longer pendulum (1.44 m)**
- **6 oscillations of the shorter pendulum (1 m)**

### Final Answer
The answer is **6 oscillations of the smaller pendulum**.

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