Question

A 19.4 cm pendulum has a period of 0.88 s. What is the free-fall acceleration at the pendulum's location?

Answer 1

Acceleration due to gravity value

`=9.89m/ {s}^(2)`

Step-by-step explanation:

Time period of simple pendulum is given by the expression

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

Here we have

T = 0.88 s

l = 19.4 cm = 0.194 m

Substituting

`0.88=2\pi\sqrt{(0.194)/(g)}\\\\(0.194)/(g)=0.0196\\\\g=9.89m/s^2 \\`

Acceleration due to gravity value

`=9.89m/s^2 \\`