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

Answer:

g = 12.22 m/s²

Explanation:

The time period of this pendulum is given as follows:

$T = \frac{time\ taken}{no. of cycles}\\\\T = \frac{144\ s}{93}\\\\T = 1.55\ s$

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

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

where,

L = length of pednulum = 48 cm = 0.48 m

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

Therefore,

$1.55\ s=2\pi \sqrt{\frac{0.48\ m}{g}}\\\\\sqrt{g} = 2\pi \sqrt{\frac{0.48\ m}{1.55\ s}}\\\\g = 3.49^{2}$

g = 12.22 m/s²