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

Answer:

The  value is  $g = 16.104 \ m/s$

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

    $T = 2 \pi \sqrt{\frac{ l}{g} }$

Also the period is mathematically represented as

      $T = \frac{ t}{n}$

      $T = \frac{125}{110}$

      $T = 1.14 \ s/cycle$

So

      $1.14 = 2 * 3.142 \sqrt{\frac{ 0.53}{g} }$

=>   $g = \frac{ 4 * 3.142^2 * 0.53}{1.14^2}$

=>     $g = 16.104 \ m/s$