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?

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!

snehashish65 snehashish65 · Answer 2 · 2021-11-19T12:40:32+01:00

(a) The value of pendulum's time constant is 53.38 minutes.

(b) The additional time elapses before the amplitude decreases to 25% of its initial value is 15.35 minutes.

Given data:

The percentage decrease in oscillation is, 50%.

The time interval is, t = 37 mins = 2220 s.

(a)

The relation for the amplitude decay is given as,

[tex]A '= Ae^{-t/T}[/tex]

Here, A' is the final amplitude and A is the initial amplitude. And T is the time constant.

Solving as,

[tex]A (1-\dfrac{50}{100} )= Ae^{-2220/T}\\\\0.5 = e^{-2220/T}\\\\ln(0.5) = -2220/T\\\\T = 3203.46 s /60 s\\\\T = 53.38 \;\rm mins[/tex]

Thus, the value of pendulum's time constant is 53.38 minutes.

(b)

Now for 25% time decay, the time is,

[tex]A (1-\dfrac{25}{100} )= Ae^{-t/T}\\\\0.75 = e^{-t/53.38}\\\\ln(0.75) = -t/53.38\\\\t = 15.35 \;\rm mins[/tex]

So, additional time elapse will be 15.35 minutes.

Thus, additional time elapses before the amplitude decreases to 25% of its initial value is 15.35 minutes.

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