Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 13 - Exercises and Problems - Page 247: 60

Answer

System B's spring constant is four times that of system A.

Work Step by Step

We know the total energy of a spring mass system $(E)=\frac{1}{2}KA^{2}$ , Where. K - spring constant, a - amplitude of the oscillation. For the system A, let's assume $K=K_{1}, A=A$ & for the system B, $K=K_{2},\space A=A/2$ & apply above equation to both systems. $E_{A}=\frac{1}{2}K_{1}A^{2}-(1)$ $E_{B}=\frac{1}{2}K_{2}(\frac{A}{2})^{2}-(2)$ (1)=(2) => $\frac{1}{2}K_{1}A^{2}=\frac{1}{2}K_{2}(\frac{A}{2})^{2}$ $K_{1}=\frac{K_{2}}{4}=>4K_{1}=K_{2}$ System B's spring constant is four times that of system A
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