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

Answer:

7.50 m/s^2

Explanation:

The period of a pendulum is given by:

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

$f=\frac{N}{t}=\frac{20}{35.5 s}=0.563 Hz$

And the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{0.563 Hz}=1.776 s$

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

$g=\frac{4 \pi^2 L}{T^2}=\frac{4 \pi^2 (0.600 m)}{(1.776 s)^2}=7.50 m/s^2$