Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 13 - Exercises and Problems - Page 247: 66

Answer

The proof is below.

Work Step by Step

Equation 13.19 states: $A(\omega)=\frac{F_d}{m\sqrt{(\omega_d^2-\omega_0^2)^2+\frac{b^2\omega_d^2}{m^2}}}$ We plug in $b= \sqrt2m\omega_0$ to find: $A(\omega)=\frac{F_d}{m\sqrt{(\omega_d^2-\omega_0^2)^2+2\omega_0^2\omega_d^2}}$ We take the derivative and set it equal to zero to find the maximum drive frequency. Doing this, we find that: $\omega_d^2 = \omega_0^2 -.5b^2 \times\frac{1}{m^2}$ Plugging in $b= \sqrt2m\omega_0$, we find that this equation simplifies to: $w_d^2 =0$ Thus, if $\omega_0$ is less than $b= \sqrt2m\omega_0$, it follows that the maximum of $\omega_d$ is less than $\omega_0$.
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