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