# Chapter 13 - Exercises and Problems - Page 238: 31

1.64 seconds

#### Work Step by Step

Substituting the value of $I=\frac{4}{3}mL^2$ into the equation, we find that the following equation gives the period of the meter stick: $T = 2\pi\sqrt{\frac{\frac{4}{3}mL^2}{mgL}}$ Recall, the length is equal to half of the length of the meter stick, which is half of a meter. Thus, we find: $T = 2\pi\sqrt{\frac{\frac{4}{3}mL^2}{mgL}}$ $T = 2\pi\sqrt{\frac{\frac{4}{3}L}{g}}$ $T = 2\pi\sqrt{\frac{\frac{4}{3}(.5)}{9.81}}=\fbox{1.64 seconds}$

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