Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

Chapter 15 - Oscillations - Exercises and Problems - Page 416: 38

Answer

(a) The fraction of energy that is potential energy is 0.25. The fraction of energy that is kinetic energy is 0.75 (b) $x = 0.707~A$

Work Step by Step

(a) We can find an expression for the total energy in the system when all the energy is in the form of elastic potential energy. $E = \frac{1}{2}kA^2$ We can find an expression for the elastic potential energy when $x = \frac{1}{2}A$ $U_s = \frac{1}{2}kx^2$ $U_s = \frac{1}{2}k~(\frac{1}{2}A)^2$ $U_s = \frac{1}{2}k~(\frac{1}{4}A^2)$ We can find the fraction of energy that is potential energy. $\frac{U_s}{E} = \frac{\frac{1}{2}k~(\frac{1}{4}A^2)}{\frac{1}{2}kA^2}$ $\frac{U_s}{E} = \frac{1}{4} = 0.25$ The fraction of energy that is potential energy is 0.25. Therefore, the fraction of energy that is kinetic energy is 0.75 (b) We can find $x$ when half the total energy is potential energy. $U_s = \frac{E}{2}$ $\frac{1}{2}kx^2 = \frac{\frac{1}{2}kA^2}{2}$ $x^2 = \frac{A^2}{2}$ $x = \sqrt{\frac{1}{2}}~A$ $x = 0.707~A$

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