osciillating on the moon that has the same period as a 4.87 m pendulum oscilating on Earth. If the moon's gravity is one-sixth of the Earth's, what is the length of the pendulum on the moon (in units of m)??
2006-11-29 12:33:08 · 5 answers · asked by Mariska 2 in Science & Mathematics ➔ Physics
The square of the period is directly proportional to the length and inversely proportional to the acceleration.
Thus, if the period is the same, then acceleration per unit length is a constant.
g/ L is a constant.
g(e) /L(e) = g(m)/L(m) ---------------(e) for earth and (m) for moon.
Lm / Le = (gm/ge) = 1/6
Lm = (1/6) * Le
Lm = (1/6) * 4.87
Lm = 0.812m.
2006-11-29 12:58:31 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
For the same period the length must be 6 x shorter. So the length must be 0.81 m.
2006-11-29 20:38:25 · answer #2 · answered by Robert O 2 · 0⤊ 0⤋
Here is a web site with the answer:
http://online.cctt.org/physicslab/content/Phy1HON/lessonnotes/pendulums/period.asp
2006-11-29 20:35:24 · answer #3 · answered by firefly 6 · 0⤊ 0⤋
.812 m
2006-11-29 20:35:39 · answer #4 · answered by joshua h 3 · 0⤊ 0⤋
ewwwwwwwwwww physics
lmaoo srry...i just felt like i had to share that
2006-11-29 20:35:12 · answer #5 · answered by noname 2 · 0⤊ 0⤋