Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 13 - Exercises and Problems - Page 247: 64

Answer

$T = \frac{2\pi}{\sqrt{\frac{2amg}{-x}}}$

Work Step by Step

We first write an equation for the gravitational potential energy of the mass: $U=mgh = amgx^2$ We know the following equation for the angular velocity: $\frac{dx^2}{dt^2}=-\omega^2 x $ Thus, we take the second derivative of the original equation to find: $2amg = -\omega^2 x $ $ \omega = \sqrt{\frac{2amg}{-x}}$ This means that the period is: $T = \frac{2\pi}{\sqrt{\frac{2amg}{-x}}}$
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