Two simple pendulum have lengths land `( 25l)/( 16)` . At ` t=0` they are in same phase after how may oscillations of smaller pendulum will they be again in phase for first time ?

Answer

To solve the problem of finding out after how many oscillations of the smaller pendulum will both pendulums be in phase again, we can follow these steps:

### Step 1: Identify the lengths of the pendulums
Let the length of the first pendulum be \( L_1 = l \) and the length of the second pendulum be \( L_2 = \frac{25l}{16} \).

### Step 2: Calculate the time periods of the pendulums
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

For the first pendulum:
\[ T_1 = 2\pi \sqrt{\frac{l}{g}} \]

For the second pendulum:
\[ T_2 = 2\pi \sqrt{\frac{25l/16}{g}} = 2\pi \sqrt{\frac{25l}{16g}} = 2\pi \cdot \frac{5}{4} \sqrt{\frac{l}{g}} = \frac{5\pi}{2} \sqrt{\frac{l}{g}} \]

### Step 3: Find the ratio of the time periods
Now, we can find the ratio of the time periods \( T_1 \) and \( T_2 \):
\[ \frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{l}{g}}}{\frac{5\pi}{2} \sqrt{\frac{l}{g}}} = \frac{2}{\frac{5}{2}} = \frac{4}{5} \]

### Step 4: Determine the least common multiple (LCM) of the time periods
Let’s denote the time period of the first pendulum as \( T_1 = 4x \) and the second pendulum as \( T_2 = 5x \) (where \( x \) is a common factor). The LCM of \( T_1 \) and \( T_2 \) can be calculated as:
\[ \text{LCM}(T_1, T_2) = \text{LCM}(4x, 5x) = 20x \]

### Step 5: Calculate the number of oscillations of the smaller pendulum
Now, we need to find how many oscillations the smaller pendulum (first pendulum) completes in the time \( 20x \):
\[ \text{Number of oscillations of the first pendulum} = \frac{\text{Total time}}{T_1} = \frac{20x}{4x} = 5 \]

### Conclusion
Thus, after 5 oscillations of the smaller pendulum, both pendulums will be in phase again for the first time.