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

**Answer 1**

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.

**Answer 2**

Bob Mass

**Answer 3**

1) B

2) C

3) C