The time period, $T$ seconds, of a simple pendulum of length $l$ cm is given by the formula $$T = 2\pi \sqrt{\frac{l}{g}}$$ Katie uses a simple pendulum in an experiment to find an estimate for the value of $g$ - Edexcel - GCSE Maths - Question 3 - 2021 - Paper 2

The time period $T = 1.45$ seconds has an upper bound of $1.45 + 0.005 = 1.455$ seconds and a lower bound of $1.45 - 0.005 = 1.445$ seconds.

Starting with the formula for $T$, we rearrange it to solve for $g$:

$g = \frac{4\pi^2 l}{T^2}$.

Using upper bounds:
Upper bound for $g$ is calculated as:

$g_{upper} = \frac{4 \times (3.142)^2 \times 52.05}{(1.455)^2} \approx 9.88 \, \text{m/s}^2.$

Using lower bounds:
Lower bound for $g$ is calculated as:

$g_{lower} = \frac{4 \times (3.142)^2 \times 51.95}{(1.445)^2} \approx 9.79 \, \text{m/s}^2.$