Question

A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant plane. If 20 oscillations are completed in 35.5 s, the acceleration would be

Answer 1

7.50 m/s^2

Step-by-step explanation:

The period of a pendulum is given by:

`T=2\pi \sqrt{(L)/(g)}` (1)

where

L = 0.600 m is the length of the pendulum

g = ? is the acceleration due to gravity

In this problem, we can find the period T. In fact, the frequency is equal to the number of oscillations per second, so:

`f=(N)/(t)=(20)/(35.5 s)=0.563 Hz`

And the period is the reciprocal of the frequency:

`T=(1)/(f)=(1)/(0.563 Hz)=1.776 s`

And by using this into eq.(1), we can find the value of g:

`g=(4 \pi^2 L)/(T^2)=(4 \pi^2 (0.600 m))/((1.776 s)^2)=7.50 m/s^2`