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Pendulum's period on the earth to period on a moon: moon's gravity acceleration.

A simple pendulum, length L, completes 12 oscillations in 30 seconds here on the earth where g= 9.8 nt/kg. The same length pendulum on a certain moon of Jupiter, completes 18 oscillations in 54 seconds.

a.) Find the acceleration of gravity, g1, on that moon.

b.) Find the weight on the surface of that moon, of a person whose weight on the surface of the earth is W=686 nt.

Solution This solution is FREE courtesy of BrainMass!

a.) From the equation for a simple pendulum executing simple harmonic motion (SHM), recall that if the length of the pendulum is L and the gravity field is g, the period of the SHM, is given by:
(1) T = (2 Pi) (sqrt[L/g])
The period of any oscillation is:
(2) The time t divided by the number of oscillations, N.
Step 1. Use (2) to determine the periods on earth and on that moon.
Step 2. Solve the general equation (1) for L, then equate L for the earth pendulum to the same L for the moon and solve for gmoon.
This should give you:
(3) gmoon = 6.8 nt/kg.

b.) The weight of mass m, in the gravity field g, is W=mg. The person would have the same mass on the moon as on the Earth. Solving for m and equating should give you:
The weight on the moon is Wmoon= 476 nt.