College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 13 - Problems - Page 497: 14


The fractional change in the period is $0.0002$

Work Step by Step

We can find the increase in length of the steel rod when it is heated: $\Delta L = \alpha~\Delta T~L$ $\Delta L = (12\times 10^{-6}~K^{-1})(-25~K)~(2.5000~m)$ $\Delta L = -7.50~\times 10^{-4}~m$ We can find ratio of the new length to the original length: $\frac{2.5000~m-7.50~\times 10^{-4}~m}{2.5000~m} = 0.9997$ Let $T$ be the original period of the pendulum. We can find an expression for the new period of the pendulum: $T' = 2\pi\sqrt{\frac{I}{mg(L'/2)}}$ $T' = 2\pi\sqrt{\frac{\frac{1}{3}mL'^2}{mg(L'/2)}}$ $T' = 2\pi\sqrt{\frac{2L'}{3g}}$ $T' = 2\pi\sqrt{\frac{2(0.9997~L)}{3g}}$ $T' = \sqrt{0.9997}\times 2\pi\sqrt{\frac{2L}{3g}}$ $T' = \sqrt{0.9997}\times T$ $T' = 0.9998\times T$ The fractional change in the period is $1-0.9998$ which is $0.0002$
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