A simple pendulum with a length 1.50 m makes 81.6 oscillations in 180 s. What is the acceleration due to gravity at its location? (It is not located on Earth!)
2007-11-09 05:29:12 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Find its period, T.
T= 180s/81.6=2.21 s
ANSWER: g= 12.12 m/s^2 ( Saturn ?)
T=2(pi) sqrt(L/g)
Solve for g (which is on another planet.)
g = L(2pi/T)^2
2007-11-09 05:40:04 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The period T of a pendulum is approximately
T â 2Ï sqrt(L/g)
where L is the length of the pendulum and g is the location's acceleration due to gravity. T is the time it takes for the pendulum to go from vertical to its highest, back to vertical and to its highest in the other direction, then back to vertical (I'm not sure if this motion is considered one "oscillation", or two).
2007-11-09 13:45:46 · answer #2 · answered by Ron W 7 · 0⤊ 0⤋
distance times time =velocity
2007-11-09 13:36:07 · answer #3 · answered by mac 2 · 0⤊ 1⤋