The period of a pendulum oscillating due to gravity can be found using the length of the pendulum $L$ as follows:

$$
begin{equation}
T = 2 , pi sqrt{frac{L}{g}}
end{equation}
$$

By looking at formula (1) we can see that the period $T$ of the pendulum is proportional to the square root of the length of the pendulum $L$, that is:

$$
begin{align*}
T sim sqrt{L}
end{align*}
$$

From this it is obvious that a decrease in the length of the pendulum $L$ will lead to a decrease in the pendulum’s period $T$.

The period of a pendulum oscillating due to gravity can be found using the length of the pendulum $L$ as follows:

$$
begin{equation}
T = 2 , pi sqrt{frac{L}{g}}
end{equation}
$$

By looking at formula (1) we can see that the period $T$ of the pendulum is independent of the mass of the bob attached to it.

From this it is obvious that a increase in the mass of the bob will NOT change the period of the pendulum.

The period of a pendulum oscillating due to gravity can be found using the length of the pendulum $L$ as follows:

$$
begin{equation}
T = 2 , pi sqrt{frac{L}{g}}
end{equation}
$$

We know that the period of the grandfather clock is $T = 2text{ s}$

We will use formula (1) to find the length of the pendulum.

First we need to rearrange it for $L$ as follows:

$$
begin{align*}
T &= 2 , pi ; sqrt{frac{L}{g}} \
T^2 &= 4 , pi^2 ; frac{L}{g} \
L &= left(frac{T}{2 , pi} right)^2 , g
end{align*}
$$

Plugging in the values we get:

$$
begin{align*}
L &= left(frac{2text{ s}}{2 , pi} right)^2 cdot 9.81 ; frac{text{m}}{text{s}^2} \
L &approx 1text{ m}
end{align*}
$$