Two simple pendulums of length 1 m and 4 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to:

The time period of a simple pendulum is given by $$ T = 2\pi \sqrt{\frac{L}{g}} $$, where $$ L $$ is the length and $$ g $$ is the acceleration due to gravity. For the shorter pendulum of length 1 m, the time period is $$ T_1 = 2\pi \sqrt{\frac{1}{g}} $$. For the longer pendulum of length 4 m, the time period is $$ T_2 = 2\pi \sqrt{\frac{4}{g}} = 2\pi \cdot \sqrt{4} \cdot \sqrt{\frac{1}{g}} = 2\pi \cdot 2 \cdot \sqrt{\frac{1}{g}} = 4\pi \sqrt{\frac{1}{g}} $$. Comparing $$ T_1 $$ and $$ T_2 $$, we see $$ T_2 = 2 \cdot (2\pi \sqrt{\frac{1}{g}}) = 2 T_1 $$.

The angular frequency $$ \omega $$ is related to the time period by $$ \omega = \frac{2\pi}{T} $$. So for the shorter pendulum, $$ \omega_1 = \frac{2\pi}{T_1} $$. For the longer pendulum, $$ \omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{2 T_1} = \frac{\pi}{T_1} $$. The difference in angular frequencies is $$ \omega_1 - \omega_2 = \frac{2\pi}{T_1} - \frac{\pi}{T_1} = \frac{\pi}{T_1} $$.

The phase difference between the two pendulums at time $$ t $$ is $$ \Delta \theta = (\omega_1 - \omega_2) t = \left( \frac{\pi}{T_1} \right) t $$. They are in phase when this phase difference is an integer multiple of $$ 2\pi $$, so $$ \Delta \theta = 2\pi n $$ for some integer $$ n $$. Substituting, we get $$ \frac{\pi}{T_1} t = 2\pi n $$. Solving for $$ t $$, divide both sides by $$ \pi $$: $$ \frac{t}{T_1} = 2n $$, so $$ t = 2n T_1 $$.

The time $$ t $$ is also the time taken by the shorter pendulum to complete $$ N $$ oscillations, which is $$ t = N T_1 $$. Setting this equal to the expression above, $$ N T_1 = 2n T_1 $$. Dividing both sides by $$ T_1 $$ (assuming $$ T_1 \neq 0 $$), we get $$ N = 2n $$. The smallest positive integer $$ N $$ occurs when $$ n = 1 $$, giving $$ N = 2 $$. This means after 2 oscillations of the shorter pendulum, they are again in phase for the first time after the initial displacement.

Verifying with the time: at $$ t = 2 T_1 $$, the shorter pendulum completes exactly 2 oscillations. The longer pendulum has a time period of $$ 2 T_1 $$, so in time $$ 2 T_1 $$, it completes exactly one oscillation ($$ \frac{t}{T_2} = \frac{2 T_1}{2 T_1} = 1 $$). Both pendulums return to their starting positions with the same velocity direction, confirming they are in phase.

Hence, the correct answer is Option A.