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 the moon. 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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