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?

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!

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