Two simple pendulum of length l and 16l are released from the same phase together. They will be at the same time phase after a minimum time.

Options

A. `(8pi)/3sqrt(l/g)`
B. `pi/3sqrt(l/g)`
C. 2s
D. None of these

Answer

To solve the problem of determining the minimum time after which two simple pendulums of lengths \( l \) and \( 16l \) will be in the same phase again, we can follow these steps:

### Step 1: Determine 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 of length \( l \):

\[ T_1 = 2\pi \sqrt{\frac{l}{g}} \]

For the second pendulum of length \( 16l \):

\[ T_2 = 2\pi \sqrt{\frac{16l}{g}} = 2\pi \sqrt{16} \sqrt{\frac{l}{g}} = 8\pi \sqrt{\frac{l}{g}} \]

### Step 2: Calculate the Angular Frequencies
The angular frequency \( \omega \) is related to the time period by:

\[ \omega = \frac{2\pi}{T} \]

For the first pendulum:

\[ \omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{2\pi \sqrt{\frac{l}{g}}} = \sqrt{\frac{g}{l}} \]

For the second pendulum:

\[ \omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{8\pi \sqrt{\frac{l}{g}}} = \frac{1}{4} \sqrt{\frac{g}{l}} \]

### Step 3: Set Up the Phase Condition
The pendulums are released from the same phase, so we need to find the time \( T \) when they are again in the same phase. The condition for being in the same phase can be expressed as:

\[ \omega_1 T = \omega_2 T + 2n\pi \]

where \( n \) is an integer. Rearranging gives:

\[ (\omega_1 - \omega_2) T = 2n\pi \]

### Step 4: Substitute the Angular Frequencies
Substituting \( \omega_1 \) and \( \omega_2 \):

\[ \left(\sqrt{\frac{g}{l}} - \frac{1}{4}\sqrt{\frac{g}{l}}\right) T = 2n\pi \]

This simplifies to:

\[ \left(\frac{3}{4}\sqrt{\frac{g}{l}}\right) T = 2n\pi \]

### Step 5: Solve for Minimum Time \( T \)
To find the minimum time, we set \( n = 1 \):

\[ \frac{3}{4}\sqrt{\frac{g}{l}} T = 2\pi \]

Solving for \( T \):

\[ T = \frac{2\pi \cdot 4}{3\sqrt{\frac{g}{l}}} = \frac{8\pi}{3}\sqrt{\frac{l}{g}} \]

### Final Result
Thus, the minimum time after which the two pendulums will be in the same phase again is:

\[ T = \frac{8\pi}{3}\sqrt{\frac{l}{g}} \]