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 - General Problems - Page 326: 80

Answer

0.69 meters.

Work Step by Step

The surface wave is moving the ground vertically. If the object is to stay in contact with the ground at all times, the normal force on the object, $F_N$, is nonzero. The normal force on the object is directed upward. Choosing upward to be the positive direction, Newton’s second law tells us that $F_N-mg = ma$, or $F_N=mg+ma$. When the ground is accelerating upward, the acceleration a is positive, and the object never loses contact with the ground. However, we see that a downward acceleration could cause the normal force to equal zero. The critical point occurs when the magnitude of the acceleration a is equal to g. If the downward acceleration were any larger, the ground would “fall” downward at a greater rate than the object. For SHM, we know the magnitude of the maximum acceleration. $$a_{max}= \omega ^2 A = (2 \pi f)^2 A=g$$ Solve for the critical amplitude, A. $$A=\frac{g}{4\pi^2f^2}=\frac{9.80\;m/s^2}{4 \pi^2(0.60\;Hz)^2}=0.69\;m$$
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