$$T = 2\pi \sqrt{\frac{L}{g}}$$
where:
- \(T\) is the period of the pendulum in seconds (s)
- \(L\) is the length of the pendulum in meters (m)
- \(g\) is the acceleration due to gravity in meters per second squared (\(\text{m}/\text{s}^2\))
We are given that:
- \(L = 45 \text{ cm} = 0.45 \text{ m}\)
- \(g = 9.81 \text{ m}/\text{s}^2\)
Substituting these values into the formula, we get:
$$T = 2\pi \sqrt{\frac{0.45 \text{ m}}{9.81 \text{ m}/\text{s}^2}} = 1.37 \text{ s}$$
Therefore, the period of a simple pendulum 45 cm long on Earth is 1.37 seconds.
