Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. The explorer finds that the pendulum completes 108 full swing cycles in a time of 143 s.

What is the magnitude of the gravitational acceleration on this planet?

Express your answer in meters per second per second.

Answer 1

If g' in the acceleration length of the pendula in then its time period gravitational I ii the T = 20 Mgi Here = l: 55 cm 0.55 m. It completes log full swing in á time 1435. Hence T: 143 108 sec. Total time Total oscillation Wah Now, T 2 2 TT Yg T2. = 411? l gi g' 4T1² l T2, 492 x 0.55 X 108 2 143 Gravitational acceleration 00 :/12-39 m/s²