Question

58) A simple pendulum of length 2.3 m makes 5.0 complete swings in 37 s. What is the acceleration of gravity at the location of the pendulum?

Answer 1

Take into account that the period of a pendulum is given by the following formula:

`T=2\pi\sqrt[]{(l)/(g)}`

where,

l: length = 2.3 m

g: acceleration of gravity = ?

T: period

The period T is:

`T=(37s)/(5.0)=7.4s`

If you solve for g the equation of the period, you obtain:

`\begin{gathered} T^2=4\pi^2(l)/(g) \\ g=(4\pi^2l)/(T^2) \\ g=(4\pi^2(2.3m))/((7.4s)^2)\approx1.66(m)/(s^2) \end{gathered}`

Hence, the acceleration of gravity at the location of the pendulum is approximately 1.66m/s^2