Aphysics 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?

answer: part b can be solved by using the equation t=(time constant)(ln2)

t=(53)(ln2)

t=(53)(0.693)

t=37 min

here is how you get the answer for the first part of the problem!

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!

explanation:

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