A pendulum oscillating on the moon has the
same period as a(n) 2.55 m pendulum oscillating
on Earth. If the moon’s gravity is one-sixth of Earth’s
gravity, find the length of the pendulum on
Answer in units of m
The equation governing pendulums is as follows:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum and g is the gravitational constant.
Since the question states that the periods are equal on both the moon and earth, we know that
2π√(L e/g e) = 2π√(L m/g m)
where g e is gravity on earth and g m is gravity on the moon. The same goes for the lengths.
This equation can be simplified by dividing both sides by 2 pi and then squaring both sides to remove the square root, Leaving:
L e / g e = L m / gm
It is then given that L e is 2.55 m and that g m = 1/6 * g e. Substituting these values into the equation above leaves:
2.55 / g e = L m / (1/6 * g e)
Cross multiply to get
2.55 *1/6 * g e = L m * g e
Gravitational constants cancel, leaving
L m = 0,43 m
as the length of the pendulum on the moon.
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