Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm. The explorer finds that the pendulum completes 105 full swing cycles in a time of 125 s.

What is the value of the acceleration of gravity on this planet?Express your answer in meters per second per second.

Answer 1

The value is
`g =  16.104 \  m/s`

Step-by-step explanation:

From the question we are told that

The length is
`l = 53.0 \ cm = 0.53 \ m`

The number of cycle is
`n = 105`

The time taken is
`t = 125 \ s`

The period is mathematically represented as

`T  =  2 \pi \sqrt{( l)/(g) }`

Also the period is mathematically represented as

`T    =  ( t)/(n)`

`T  =  (125)/(110)`

`T  =  1.14 \ s/cycle`

So

`1.14  =  2 * 3.142  \sqrt{( 0.53)/(g) }`

=>
`g =  ( 4 *  3.142^2 *  0.53)/(1.14^2)`

=>
`g =  16.104 \  m/s`