Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 48.0 cm . The explorer finds that the pendulum completes 93.0 full swing cycles in a time of 144 s . Part A What is the magnitude of the gravitational acceleration on this planet

Answer 1

g = 12.22 m/s²

Step-by-step explanation:

The time period of this pendulum is given as follows:

`T = (time\ taken)/(no. of cycles)\\\\T = (144\ s)/(93)\\\\T = 1.55\ s`

but the formula for the time period of a simple pendulum is as follows:

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

where,

L = length of pednulum = 48 cm = 0.48 m

g = magnitude of th gravitational acceleration on this planet = ?

Therefore,

`1.55\ s=2\pi \sqrt{(0.48\ m)/(g)}\\\\√(g) = 2\pi \sqrt{(0.48\ m)/(1.55\ s)}\\\\g = 3.49^(2)\\`

g = 12.22 m/s²