Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 11 - Oscillations and Waves - Problems - Page 324: 33

Answer

One-third of the time.

Work Step by Step

Suppose the pendulum starts at maximum displacement. The equation of motion is known. $$\theta=\theta_o cos \omega t=\theta_o cos \frac{2 \pi t}{T}$$ Find the time at which the position is half the amplitude. $$0.5\theta_o =\theta_o cos \frac{2 \pi t}{T}$$ $$cos^{-1}0.5 = \frac{2 \pi t}{T}$$ $$\frac{\pi}{3} = \frac{2 \pi t}{T}$$ $$t = \frac{T}{6}$$ This is the time it takes for the pendulum to swing from 10 degrees to 5 degrees. We know it takes $\frac{T}{4}$ for the pendulum to swing from 10 degrees to 5 degrees. Thus it takes the difference, or $\frac{T}{12}$, for the pendulum to swing from 5 degrees to zero. During this first quarter-cycle, the fraction of time spent between 0 and 5 degrees is $\frac{T/12}{T/4}=\frac{1}{3}$. Due to the symmetry of motion, the behavior during each of the other quarter-cycles is the same. The pendulum is between -5 and 5 degrees one-third of the time.
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