Question

The length of a displaced pendulum body which passes its lowest point twice every second

Answer 1

Answer:

0.248 m

Explanation:

The period of a simple pendulum is given by

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

where

L is the length of the pendulum

g is the acceleration of gravity

The pendulum in the problem passes its lowest point twice every second: this means: this means that it makes one complete oscillation in one second, so its period is 1 second:

T = 1 s

And using

$g=9.8 m/s^2$

We can rearrange the equation above to find L, the length of the pendulum:

$L=\frac{T^2 g}{4\pi^2}=\frac{1^2(9.8)}{4\pi^2}=0.248 m$