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 $\omega_0$.
