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Question

the moon, the acceleration due to gravity is one-sixth that of earth. That is

22 gmoon = gearth /6 = (9.8 m/s )/6 = 1.63 m/s .

What effect, if any, would this have on the period of a pendulum of length L? How would the period of this pendulum differ from an equivalent one on earth?

Answer 1

The formula for the period of a pendulum (small amplitude of motion) is

T = 2 π sqrt( L/g ) where L is the length of the simple pendulum.

Since the gmoom is 1/6 gearth, Tmoon = Tearth sqrt(6)

Answer 2

To answer this question, consider that on the earth,

2π/T = √(g/L) ⇒

T = 2π √(L/g)

T = period

g = acceleration of gravity

L = length of pendulum

Assuming L is the same on the earth and the moon, simply replace g with g/6 in the equation for T.

T₂ = 2π √[L/(g/6)] = √6 [2π √(L/g)] = √6 T

T₂ = √6 T

Now compare T₂ to T. What changed? There is your answer.

Answer 3