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

Published by Pearson

# Chapter 5 - Force and Motion - Exercises and Problems: 11

#### Answer

The period of the 3.0-m long pendulum is 3.67 seconds.

#### Work Step by Step

Let $T$ be the period of a pendulum and let $L$ be the length of a pendulum. Therefore; $T \propto \sqrt{L}$ $T_1 \propto \sqrt{L_1}$ $T_2 \propto \sqrt{L_2}$ We can use the ratio of $\frac{T_2}{T_1}$ to find $T_2$. $\frac{T_2}{T_1} = \frac{\sqrt{L_2}}{\sqrt{L_1}}$ $T_2 = T_1~\frac{\sqrt{L_2}}{\sqrt{L_1}}$ $T_2 = (3.0~s)~\frac{\sqrt{3.0~m}}{\sqrt{2.0~m}}$ $T_2 = 3.67~s$ The period of the 3.0-m long pendulum is 3.67 seconds.

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