Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm. The explorer finds that the pendulum completes 99.0 full swing cycles in a time of 125 s. What is the value of the acceleration of gravity on this planet?

Answer 1

"12.122 m/s²" is the appropriate solution.

Step-by-step explanation:

The given values in the question are:

Length,

l = 49.0 cm

or,

=
`49.0* 10^(-2) \ m`

Time taken,

`T = (125)/(99.0)`

`=1.2626 \ s`

Now,

As we know,

⇒
`T = 2 \pi\sqrt{(l)/(g) }`

or,

⇒
`T^2=4 \pi^2[(l)/(g) ]`

`g = 4\pi^2[(l)/(T^2) ]`

By substituting the above given values, we get

`=(4* (3.14)^2* (49.0* 10^(-2)))/((1.2626)^2)`

`=(39.4384* 49.0* 10^(-2))/(1.59415876)`

`=(19.324816)/(1.59415876)`

`=12.122 \ m/s^2`