### Theoretical reminder

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

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

For a body moving with acceleration $a$ and starting from rest we know that the total traveled distance $S$ during time $t$ is given by:

$$
begin{equation}
S = frac{1}{2} cdot a cdot t^2
end{equation}
$$

section*{Calculation}
begin{enumerate}[a)]
item
Firstly we need to find the intensity of the gravitational acceleration acting on the body. Since it falls $S = 1text{ m}$ in $t = 0.451text{ s}$ we use formula (2) to write:
begin{align*}
g = 2 cdot frac{S}{t^2}
end{align*}
Plugging in the numbers we get:
begin{align*}
g = 2 cdot frac{1text{ m}}{(0.451text{ s})^2} = 9.833 ; frac{text{m}}{text{s}^2}
end{align*}
This is the required local gravitational acceleration.
item
the period of a pendulum of length $L = 0.5text{ m}$ at that location can be found using formula (1) and the result of part a as follows:
begin{align*}
T = 2 , pi ; sqrt{frac{L}{g}}
end{align*}
Plugging in the numbers we get:
begin{align*}
T &= 2 , pi sqrt{frac{0.5text{ m}}{9.833 ; frac{text{m}}{text{s}^2}}} \
T &= 1.415text{ s}
end{align*}
end{enumerate}

begin{enumerate}[a)]
item
$g = 9.833 ; frac{text{m}}{text{s}^2}$
item
$T = 1.415text{ s}$
end{enumerate}

### Theoretical reminder

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 period $T$ is defined as the time required to complete one full cycle of a given motion.
The frequency $f$ is the number of oscillations per unit time.

They are related by the following formula:

$$
begin{equation}
f = frac{1}{T}
end{equation}
$$

### Calculation

Firstly we need to find the period of a pendulum of length $L = 1.25text{ m}$ at a location where the gravitational acceleration is $g = 9.82 ; frac{text{m}}{text{s}^2}$. We do this using formula (1):

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

Plugging in the numbers we get:

$$
begin{align*}
T &= 2 , pi sqrt{frac{1.25text{ m}}{9.82 ; frac{text{m}}{text{s}^2}}} \
T &= 2.24text{ s}
end{align*}
$$

Now we find the frequency using formula (2) as follows:

$$
begin{align*}
f = frac{1}{T} = frac{1}{2.24text{ s}} = 0.446text{ Hz}
end{align*}
$$

The frequency is $f = 0.446text{ Hz}$

### Theoretical reminder

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

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

### Analysis

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

$$
begin{align*}
T sim sqrt{frac{1}{g}}
end{align*}
$$

From this it is obvious that the period will be shorter in a place where the gravitational acceleration is larger.

Since the gravitational acceleration is greater at place A than it is at place B we deduce that the period will be shorter at A than it is at B.

The period of the pendulum at place A is smaller than it is at place B, $T_{text{A}} < T_{text{B}}$