Two pendulum of lengths `1m` and `16m` are in phase at the mean position at a certain instant of time. If T is the time period of the shorter pendulum, then the minimum time after which they will again be in phase is

Options

A. `T/3`
B. `(2T)/(3)`
C. `(4T)/(3)`
D. `(8T)/(3)`

Explanation

To solve the problem of two pendulums of lengths 1m and 16m being in phase at the mean position, we need to find the minimum time after which they will again be in phase. Here’s a step-by-step solution:

### Step 1: Calculate 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 (approximately \( 9.81 \, \text{m/s}^2 \)).

For the shorter pendulum (length \( L_1 = 1 \, \text{m} \)):
\[ T_1 = 2\pi \sqrt{\frac{1}{g}} = T \]

For the longer pendulum (length \( L_2 = 16 \, \text{m} \)):
\[ T_2 = 2\pi \sqrt{\frac{16}{g}} = 2\pi \sqrt{16} \cdot \frac{1}{\sqrt{g}} = 4T \]

### Step 2: Determine the Phase of Each Pendulum
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]

For the first pendulum:
\[ \text{Phase}_1 = \omega_1 t = \frac{2\pi}{T} t \]

For the second pendulum:
\[ \text{Phase}_2 = \omega_2 t = \frac{2\pi}{4T} t = \frac{\pi}{2T} t \]

### Step 3: Find the Phase Difference
The phase difference \( \Delta \phi \) between the two pendulums is:
\[ \Delta \phi = \text{Phase}_1 - \text{Phase}_2 = \frac{2\pi}{T} t - \frac{\pi}{2T} t \]
\[ \Delta \phi = \left( \frac{2\pi}{T} - \frac{\pi}{2T} \right) t = \left( \frac{4\pi}{2T} - \frac{\pi}{2T} \right) t = \frac{3\pi}{2T} t \]

### Step 4: Condition for Being in Phase
For the two pendulums to be in phase again, the phase difference must be an integer multiple of \( 2\pi \):
\[ \Delta \phi = 2n\pi \]
Setting \( n = 1 \) for the minimum time:
\[ \frac{3\pi}{2T} t = 2\pi \]

### Step 5: Solve for Time \( t \)
Now, we can solve for \( t \):
\[ t = \frac{2\pi \cdot 2T}{3\pi} = \frac{4T}{3} \]

### Conclusion
The minimum time after which both pendulums will again be in phase is:
\[ \boxed{\frac{4T}{3}} \]