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**Hint:**A simple pendulum exhibits a simple harmonic motion. The properties of this particular type of motion can be used to find the mechanics that any oscillating pendulum experiences.

**Formula used:**$y = a\sin \omega t$ where $y$ is the displacement in a simple harmonic motion, $a$ is the amplitude,$\omega $ is the angular frequency, and $t$ is the time.

**Complete step by step answer:**

We are aware that an oscillating pendulum undergoes simple harmonic motion (SHM). In this question, we are asked to find the speed of such a pendulum with the following specifications:

Amplitude: $a = 10cm = 0.1m$$[\because 1m = 100cm]$

Time period: $T = 4s$

Time passed since the start of motion $t = 1s$

We must remember to convert the unit into SI. We know that the displacement for SHM is given by:

$y = a\sin \omega t$

To find the velocity, we differentiate the above equation as:

$\dfrac{{dy}}{{dt}} = a\dfrac{d}{{dt}}\left( {\sin \omega t} \right)$ [$a$ comes outside as it is a constant]

\[v = \dfrac{{dy}}{{dt}} = a\omega \left( {\cos \omega t} \right)\]\[[\dfrac{d}{{dt}}\left( {\sin \theta } \right) = \cos \theta ]\]

Notice that this expression has the term $\omega $ which is the angular frequency but in the question we are given with the linear time period. Hence, we convert $T$ as:

$\omega = \dfrac{{2\pi }}{T}$

We put this expression directly in our velocity to make the calculations easier:

$v = a \times \dfrac{{2\pi }}{T} \times \cos \left( {\dfrac{{2\pi }}{T}t} \right)$

Now, we substitute the values and calculate $v$ as:

$v = 0.1 \times \dfrac{{2\pi }}{4} \times \cos \left( {\dfrac{{2\pi }}{4} \times 1} \right)$

We know that

$\cos \left( {\dfrac{{2\pi }}{4}} \right) = \cos \left( {\dfrac{\pi }{2}} \right) = 0$

This makes the value of $v = 0$.

**$\therefore $ The speed after 1 second is option A i.e. zero.**

**Note:**The time period for a simple pendulum is defined as the time taken by it to complete one to and fro motion about its mean position. While the angular frequency measures the angular displacement per unit time. Hence, a factor of $2\pi $ needs to be multiplied.

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