Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 10 - Simple Harmonic Motion and Elasticity - Problems - Page 275: 20

Answer

$k_2=696N/m$

Work Step by Step

1) The spring constant of spring 2 can be found, using the equation $$\omega=\sqrt{\frac{k}{m}}$$ $$m=\frac{k}{\omega^2}$$ We know 2 objects have the same mass, so $$\frac{k_1}{\omega^2_1}=\frac{k_2}{\omega^2_2}$$ $$k_2=k_1\frac{\omega_2^2}{\omega_1^2} (1)$$ We need to find $\frac{\omega_2}{\omega_1}$ 2) Both objects have the same magnitude of maximum velocity. Therefore, $$A_1\omega_1=A_2\omega_2$$ $$\frac{\omega_2}{\omega_1}=\frac{A_1}{A_2}=2$$ $$\frac{\omega_2^2}{\omega_1^2}=4$$ Plug this back to (1): $k_2=4k_1=696N/m$
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