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 325: 65

Answer

$A=\frac{Mg}{k}$.

Work Step by Step

If the penny to stay in contact with the block at all times, the normal force on the penny, $F_N$, is nonzero. The normal force on the penny 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 block is accelerating upward, the acceleration a is positive, and the penny never loses contact with the block. 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. For SHM, we know the magnitude of the maximum acceleration. We also know that the penny's mass, m, is much, much less than the block's mass, M. $$a_{max}=\omega^2 A=\frac{k}{M+m}A\approx\frac{k}{M}A$$ Equate the magnitude of the maximum acceleration to g, and solve. $$ \frac{k}{M}A=g$$ $$ A=\frac{Mg}{k} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.