After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 136 s. What is the value of g on this planet?

g = 10.67 m/s²

Explanation:

The period of a simple pendulum is given as;

T = 2π√(L / g)

Where;

T = period

L = length of spring

g = acceleration due to gravity

Now, in the question we want to find g and we are given L but not given the period.

We are given

So let's find the period (T)

We are given total number of oscillations and time taken to complete those oscillations. Thus, we can express period as;

T = time taken/total number of oscillations

So, T = 136/100 = 1.36

Thus,

1.36 = 2π√(L / g)

1.36/2π = √(L / g)

Taking the square of both sides, we have;

(1.36/2π)² = L/g

0.04685 = L/g

We are given that; L = 50cm = 0.5m

So,

0.04685 = 0.5/g

g = 0.5/0.04685

g = 10.67 m/s²