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 when two simple pendulums of lengths \( l \) and \( 16l \) will be at the same phase after being released from the same phase, we can follow these steps:

### Step 1: Determine the time period of each pendulum
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 pendulum of length \( l \):
\[ T_1 = 2\pi \sqrt{\frac{l}{g}} \]

- For the pendulum of length \( 16l \):
\[ T_2 = 2\pi \sqrt{\frac{16l}{g}} = 2\pi \sqrt{16} \sqrt{\frac{l}{g}} = 4 \cdot 2\pi \sqrt{\frac{l}{g}} = 4T_1 \]

### Step 2: Find the angular frequencies
The angular frequency \( \omega \) is related to the time period by:
\[ \omega = \frac{2\pi}{T} \]

- For the first pendulum:
\[ \omega_1 = \frac{2\pi}{T_1} \]

- For the second pendulum:
\[ \omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{4T_1} = \frac{1}{2} \cdot \frac{2\pi}{T_1} = \frac{1}{2} \omega_1 \]

### Step 3: Determine the phase difference
Since both pendulums are released from the same phase, we want to find the time \( t \) when their phase difference is an integer multiple of \( 2\pi \):
\[ \text{Phase difference} = \omega_2 t - \omega_1 t = (\omega_2 - \omega_1)t \]

Setting the phase difference equal to \( 2\pi n \) (where \( n \) is an integer), we have:
\[ (\omega_2 - \omega_1)t = 2\pi n \]

### Step 4: Substitute the angular frequencies
Substituting \( \omega_2 \) and \( \omega_1 \):
\[ \left(\frac{1}{2} \omega_1 - \omega_1\right)t = 2\pi n \]
This simplifies to:
\[ -\frac{1}{2} \omega_1 t = 2\pi n \]
or
\[ \frac{1}{2} \cdot \frac{2\pi}{T_1} t = 2\pi n \]

### Step 5: Solve for time \( t \)
Cancelling \( 2\pi \) from both sides:
\[ \frac{1}{2T_1} t = n \]
Thus,
\[ t = 2nT_1 \]

### Step 6: Find the minimum time
To find the minimum time when they are in the same phase, we take \( n = 1 \):
\[ t = 2T_1 \]

### Step 7: Substitute \( T_1 \)
Now substituting \( T_1 \):
\[ T_1 = 2\pi \sqrt{\frac{l}{g}} \]
Thus,
\[ t = 2 \cdot 2\pi \sqrt{\frac{l}{g}} = 4\pi \sqrt{\frac{l}{g}} \]

### Final Result
The minimum time after which both pendulums will be at the same phase is:
\[ t = 4\pi \sqrt{\frac{l}{g}} \]