A physics department has a Foucault pendulum, a long-period pendulum suspended from the ceiling. The pendulum has an electric circuit that keeps it oscillating with a constant amplitude. When the circuit is turned off, the oscillation amplitude decreases by 50% in 37 minutes.

a. What is the pendulum's time constant?

b. How much additional time elapses before the amplitude decreases to 25% of its initial value?

Question

Aden Mercer

Physics

2 August, 11:07

A physics department has a Foucault pendulum, a long-period pendulum suspended from the ceiling. The pendulum has an electric circuit that keeps it oscillating with a constant amplitude. When the circuit is turned off, the oscillation amplitude decreases by 50% in 37 minutes.

a. What is the pendulum's time constant?

b. How much additional time elapses before the amplitude decreases to 25% of its initial value?

+5

Answers (1)

Know the Answer?

Elaine Boyle · Answer 1

t=37 mins - > 2220sec

We want "T" which is the pendulum time constant

Using this equation

.5A=Ae^ (-t/T)

The. 5A is half the amplitude

Take ln of both sides to get ride of Ae

=ln (.5) = -2220/T

Now rearrange to = T

T=-2220/ln (.5) = 3202.78sec / 60 secs = 53.38 mins - > first part of the answer.

The second part is really easy. It took 37 mins to decay half way. meaning to decay another half of 50% which equals 25% it will take an additional 37 mins!