Question

A pendulum on Earth has a period of 1.2 seconds. The same pendulum on Mercury (dark side) has a period of 1.95 seconds. What is the free-fall acceleration of Mercury (in m/s^2)

Answer 1

`g'=3.71\ m/s^2`

Step-by-step explanation:

Given that,

Time period of a pendulum on the earth's surface, T₁ = 1.2 s

Time period of the same pendulum on Mercury, T₂ = 1.95 s

The time period of the pendulum is given by :

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

On earth :

`T_1=2\pi \sqrt{(l)/(g)}`

`1.2=2\pi \sqrt{(l)/(9.8)}`.............(1)

Let g' is the acceleration due to gravity on Mercury. So,

`1.95=2\pi \sqrt{(l)/(g')}`............(2)

From equation (1) and (2) :

`(1.2)/(1.95)=\sqrt{(g')/(9.8)}`

`g'=((1.2)/(1.95))^2* 9.8`

`g'=3.71\ m/s^2`

So, the acceleration due to gravity on the mercury is
`3.71\ m/s^2`. Hence, this is the required solution.