Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 15 - Oscillations - Problems: 2b

Answer

$k=130N/m$

Work Step by Step

Recall that the angular frequency of a mass-spring system is equal to $$\omega=\sqrt{\frac{k}{m}}$$ The formula relating period and angular frequency is $$T=\frac{2\pi}{\omega}$$ Substituting this expression into the period equation yields $$T=2\pi \sqrt{\frac{m}{k}}$$ Solve for $k$ to get $$\frac{T}{2\pi}=\sqrt{\frac{m}{k}}$$ $$\frac{T^2}{4\pi^2}=\frac{m}{k}$$ $$k=\frac{4\pi^2m}{T^2}$$ Substituting known values of $m=0.12kg$ and $T=0.20s$ yields a spring constant of $$k=\frac{4\pi^2(0.13kg)}{(0.20s)^2}$$ $$k=130N/m$$
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