Question

a simple pendulum of length 2.3 m makes 9.0 complete swings in 37.0 s.

part a
what is the acceleration of gravity at the location of the pendulum?
express your answer using two significant figures.

Explanation:

Calculate the period of the pendulum

Using the Simple Harmonic Motion Period knowledge point

$$ T = \frac{\text{Total time}}{\text{Number of swings}} = \frac{37.0\text{ s}}{9.0} \approx 4.111\text{ s} $$

Relate period to acceleration of gravity

The period of a simple pendulum is given by:

$$ T = 2\pi \sqrt{\frac{L}{g}} $$

Rearranging this formula to solve for \(g\):

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

Calculate the acceleration of gravity

Substitute the known values \(L = 2.3\text{ m}\) and \(T \approx 4.111\text{ s}\):

$$ g = \frac{4\pi^2 (2.3)}{(4.111)^2} \approx \frac{90.799}{16.901} \approx 5.372\text{ m/s}^2 $$

Round to two significant figures

Rounding \(5.372\text{ m/s}^2\) to two significant figures yields:

$$ g \approx 5.4\text{ m/s}^2 $$

Answer:

What is the acceleration of gravity at the location of the pendulum?

\(g =\) <blank>\(5.4\)</blank> \(\text{m/s}^2\)