Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 10 - Simple Harmonic Motion and Elasticity - Problems - Page 277: 48



Work Step by Step

1) For pendulum A, we have $$2\pi f=\frac{2\pi}{T}=\sqrt{\frac{mgL}{I}}$$ $$T=\frac{2\pi}{\sqrt{\frac{mgL}{I}}}=\sqrt{\frac{4\pi^2I}{mgL}}$$ Pendulum A is a thin rod, whose rotation axis is at one end, so $I_A=\frac{1}{3}mL^2$. The rod's center of gravity is at its center, since the rod is uniform. Therefore, $L=d/2$. $$T_A=\sqrt{\frac{4\pi^2mL^2}{3mgL}}=\sqrt{\frac{4\pi^2L}{3g}}$$ $$T_A=\sqrt{\frac{4\pi^2(d/2)}{3g}}=\sqrt{\frac{2\pi^2d}{3g}}$$ 2) Pendulum B is a simple pendulum, so $$2\pi f = \frac{2\pi}{T}=\sqrt{\frac{g}{L}}$$ $$T=\sqrt{\frac{4\pi^2L}{g}}$$ The pendulum's length $L=d$. $$T_B=\sqrt{\frac{4\pi^2d}{g}}$$ Therefore, $$\frac{T_A}{T_B}=\sqrt{\frac{\frac{2\pi^2L}{3g}}{\frac{4\pi^2L}{g}}}=\sqrt{\frac{\frac{2}{3}}{1}}=\sqrt{\frac{2}{3}}=0.82$$
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