If the bob's mass is doubled, approximately what will the pendulum's new period be?

Question

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T.

If the bob's mass is doubled, approximately what will the pendulum's new period be?

A: T/2

B: T

C: sqrt(2)*T

D: 2T

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If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now be?

A: T/6

B: T/sqrt(6)

C: sqrt(6)*T

D: 6T

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If the pendulum is taken into the orbiting space station what will happen to the bob?

A: It will continue to oscillate in a vertical plane with the same period.

B: It will no longer oscillate because there is no gravity in space.

C: It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

D: It will oscillate much faster with a period that approaches zero.

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Please choose from A-D for the 3 questions and explain your reasoning.

Answer 1

1) B

2) C

3) C

Answer 2

Pendulum period in seconds

T ≈ 2π√(L/G)

L is length of pendulum in meters

G is gravitational acceleration = 9.8 m/s²

If the bob's mass is doubled, approximately what will the pendulum's new period be?

B: T. Rate of a pendulum does not depend on the mass

If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now be?

C: sqrt(6)*T

T1 ≈ 2π√(L/G)

√(L/G) = T1/2π

(L/G) = (T1/2π)²

L = G(T1/2π)²

T2 ≈ 2π√(L/G/6)

T2 ≈ 2π√(G(T1/2π)²/G/6)

T2 ≈ 2π√((T1/2π)²/1/6)

T2 ≈ 2π√6(T1/2π)²

T2 ≈ 2π(T1/2π)√6

T2 ≈ T1√6

If the pendulum is taken into the orbiting space station what will happen to the bob?

C: It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.