Question 2.2.5: Derive equation (2.40) for the normalized magnitude and calc......

From equation (2.36) the magnitude of the steady-state response is

x_p(t)=\overbrace{\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)^2+\left(2 \zeta \omega_n \omega\right)^2}}}^X \cos (\omega t-\overbrace{\tan ^{-1} \frac{2 \zeta \omega_n \omega}{\omega_n^2-\omega^2}}^\theta)     (2.36)

X=\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)^2+\left(2 \zeta \omega_n \omega\right)^2}}=\frac{F_0 / m}{\sqrt{\left(\omega_n^2-\omega^2\right)+\left(2 \zeta \omega_n \omega\right)^2}}

Factoring \omega_n^2 out of the denominator and recalling that \omega_n^2=k / m yields

X=\frac{F_0 / m}{\omega_n^2 \sqrt{\left[1-\left(\frac{\omega}{\omega_n}\right)^2\right]^2+\left[2 \zeta\left(\frac{\omega}{\omega_n}\right)\right]^2}}=\frac{F_0 / k}{\sqrt{\left(1-r^2\right)^2+(2 \zeta r)^2}}

where r=\omega / \omega_n. Dividing both sides by F_0 / k yields equation (2.40). The maximum value of X will occur where the first derivative of X / F_0 vanishes, that is

\frac{d}{d r}\left(\frac{X k}{F_0}\right)=\frac{d}{d r}\left\{\left[\left(1-r^2\right)^2+(2 \zeta r)^2\right]^{-1 / 2}\right\}=0

Thus

r_{\text {peak }}=\sqrt{1-2 \zeta^2}=\frac{\omega_p}{\omega_n}     (2.41)

defines the value of the driving frequency, \omega_p at which the peak value of the magnitude occurs. This holds only for underdamped systems for which \zeta<1 / \sqrt{2}.

Otherwise, the magnitude does not have a maximum value or peak for any value of \omega>0 because \sqrt{1-2 \zeta^2} becomes an imaginary number for values of \zeta larger than 1 / \sqrt{2}. Note also that this peak occurs a little to the left of, or before, resonance (r=1) since

r_{\text {peak }}=\sqrt{1-2 \zeta^2}<1

This can be seen in both Figures 2.8 and 2.9. The value of the magnitude at r_{\text {peak }} is

\frac{X k}{F_0}=\frac{1}{2 \zeta \sqrt{1-\zeta^2}}     (2.42)

which is obtained simply by substituting r_{\text {peak }}=\sqrt{1-2 \zeta^2} into the expression for the normalized magnitude X k / F_0.