College Physics (4th Edition)

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

Chapter 10 - Problems - Page 401: 72


$\frac{L_2}{L_1} = 1.11$

Work Step by Step

We can find an expression for $L_1$: $T_1 = 2\pi~\sqrt{\frac{L_1}{g}}$ $(\frac{T_1}{2\pi})^2 = \frac{L_1}{g}$ $L_1 = \frac{T_1^2~g}{(2\pi)^2}$ We can find an expression for $L_2$: $T_2 = 2\pi~\sqrt{\frac{L_2}{g}}$ $(\frac{T_2}{2\pi})^2 = \frac{L_2}{g}$ $L_2 = \frac{T_2^2~g}{(2\pi)^2}$ We can find the value of the ratio $\frac{L_2}{L_1}$: $\frac{L_2}{L_1} = \frac{\frac{T_2^2~g}{(2\pi)^2}}{\frac{T_1^2~g}{(2\pi)^2}}$ $\frac{L_2}{L_1} = \frac{T_2^2}{T_1^2}$ $\frac{L_2}{L_1} = \frac{(1.00~s)^2}{(0.950~s)^2}$ $\frac{L_2}{L_1} = 1.11$
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