Question

A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant planet. If 20 complete oscillation are completed and 35.5 seconds, the acceleration is

a) 7.52 m/s squared

b) 9.81 m/s squared

c) 30.5 m/s squared

d) 42.0 m/s squared

Answer 1

Explanation :

It is given that,

length of pendulum, l = 0.6 m

Number of oscillation, n = 20

time, t = 35.5 seconds

so, Time period,
`T=(t)/(n)`

`T=(35.5s)/(20)`

we know that the time period of pendulum is given by

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

`g=(4\pi^2l)/(T^2)`

`g=(4*(3.14)^2* 0.6 * (20)^2)/((35.5s)^2)`

`g=7.51\ m/s^2`

or

`g=7.52\ m/s^2`