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 101 full swing cycles in a time of 140 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: g = 10.0 m/s/s

Explanation:

For a simple pendulum, provided that the angle between the lowest and highest point of  his trajectory be small, the oscillation period is given by the following expression:

T = 2π √L/g , where L = pendulum length, g= accelleration of gravity.

We can also define the period, as the time needed to complete a full swing, so from the measured  values, we can conclude the following :

T = 140 sec/ 101 cycles = 1.39 sec

Equating both definitions for T, we can solve for g, as follows:

g = 4 π² L / T² = 4π². 0.49 m / (1.39)² = 10.0 m/s/s