## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 14 - Oscillations - Conceptual Questions: 2

#### Answer

(a) $T' = 2.0~s$ (b) $T' = 2.83~s$ (c) $T' = 2.0~s$

#### Work Step by Step

We can write an expression for the period $T$ of an oscillating pendulum. $T = 2\pi~\sqrt{\frac{L}{g}} = 2.0~s$ (a) Since the period does not depend on the mass, the period of $T = 2.0~s$ remains the same. (b) We can find the period $T'$ when the length is doubled. $T' = 2\pi~\sqrt{\frac{2L}{g}}$ $T' = \sqrt{2}\times 2\pi~\sqrt{\frac{L}{g}}$ $T' = \sqrt{2}\times T$ $T' = (\sqrt{2})(2.0~s)$ $T' = 2.83~s$ (c) Since the period does not depend on the oscillation amplitude, the period of $T = 2.0~s$ remains the same.

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