Question

If a fixed length simple pendulum is found to have three times the period on an unknown planet’s surface (compared to Earth), what is the acceleration due to gravity on that planet? Show your work.

Answer 1

Answer:

g/9

Explanation:

length of the pendulum = L

time period on the earth = T

Time period on the planet = 3T

Let the acceleration due to gravity on the earth is g and on the planet is g'.

Use the formula for the time period of a simple pendulum for the time period on earth

$T=2\pi \sqrt{\frac{L}{g}}$     .... (1)

Time period on the surface of planet is

$3T=2\pi \sqrt{\frac{L}{g'}}$      .... (2)

Divide equation (2) by equation (1)

$\frac{3T}{T}= \sqrt{\frac{g}{g'}}$

g' = g/9

Thus, the acceleration due to gravity on the planet is g /9