Question

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

Answer 1

0.248 m

Step-by-step explanation:

The period of a simple pendulum is given by

`T=2\pi \sqrt{(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=(T^2 g)/(4\pi^2)=(1^2(9.8))/(4\pi^2)=0.248 m`