Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 15 - Oscillations - Exercises and Problems: 14

Answer

(a) $T' = 2.83~s$ (b) $T' = 1.41~s$ (c) $T' = 2.00~s$ (d) $T' = 1.41~s$

Work Step by Step

We can write an expression for the period $T$ of a mass oscillating on a spring; $T = 2\pi~\sqrt{\frac{m}{k}} = 2.00~s$ (a) We can find the period $T'$ when the mass is doubled as: $T' = 2\pi~\sqrt{\frac{2m}{k}}$ $T' = \sqrt{2}\times 2\pi~\sqrt{\frac{m}{k}}$ $T' = \sqrt{2}\times T$ $T' = (\sqrt{2})(2.00~s)$ $T' = 2.83~s$ (b) We can find the period $T'$ when the mass is halved as: $T' = 2\pi~\sqrt{\frac{m/2}{k}}$ $T' = \sqrt{\frac{1}{2}}\times 2\pi~\sqrt{\frac{m}{k}}$ $T' = \sqrt{\frac{1}{2}}\times T$ $T' = (\sqrt{\frac{1}{2}})(2.00~s)$ $T' = 1.41~s$ (c) Since the period does not depend on the oscillation amplitude, the period of $T = 2.00~s$ remains the same. (d) We can find the period $T'$ when the spring constant is doubled; $T' = 2\pi~\sqrt{\frac{m}{2k}}$ $T' = \sqrt{\frac{1}{2}}\times 2\pi~\sqrt{\frac{m}{k}}$ $T' = \sqrt{\frac{1}{2}}\times T$ $T' = (\sqrt{\frac{1}{2}})(2.00~s)$ $T' = 1.41~s$
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