T = 330/100 = 3.30 s. = The Period.
T^2 = 4*pi^2(L/g) = 3.3^2 = 10.89
39.5(1.2/g) = 10.89
47.4/g = 10.89
10.89g = 47.4
g = 4.35 m/s^2
In Concept Simulation 10.2 you can explore the concepts that are important in this problem. Astronauts on a distant planet set up a simple pendulum of length 1.2 m. The pendulum executes simple harmonic motion and makes 100 complete oscillations in 330 s. What is the magnitude of the acceleration due to gravity on this planet?
2 answers
T=330/100=3.3 s
330/100=2pi sqrt (1.2m/g)
rearrange to get :
sqrt G= (pi^2 x sqrt 1.2 )/ (3.3)
(Only square the top part of the fraction to get rid of the sqrt)
G=pi^2 x 1.2 / 3.3
g=3.588 m/s ^2
330/100=2pi sqrt (1.2m/g)
rearrange to get :
sqrt G= (pi^2 x sqrt 1.2 )/ (3.3)
(Only square the top part of the fraction to get rid of the sqrt)
G=pi^2 x 1.2 / 3.3
g=3.588 m/s ^2
1 answer