For $\frac{x}{(\frac{m_0e}{m})}$ to be maximum, the denominator in the above equation should be minimum. Thus, $\frac{d}{dr} [ ( \frac{1}{r^2} – 1)^2 \ + \ (\frac{2 \xi}{r})^2] \ = \ 0$

$2 (\frac{1}{r^2} \ – \ 1) (\frac{-2}{r^3}) \ + \ 4 \xi^2 (\frac{-2}{r^3}) \ = \ 0$

$(\frac{1}{r^2} \ – \ 1) \ + \ 2 \xi^2 \ = \ 0$

$\frac{1}{r^2} \ = \ 1 \ – \ 2 \xi^2$

$r^2 \ = \ \frac{1}{1 \ – \ 2 \xi^2}$

$r \ = \ \frac{1}{\sqrt{ 1 \ – \ 2 \xi^2}}$

The equation for phase angle remains the same as

$\phi \ = \ tan^{-1} [ \frac{ 2 \xi \frac{w}{w_n}}{ 1 \ – \ (\frac{w}{w_n})^2}]$

The variation of $\frac{x}{(\frac{m_0e}{m})}$ versus r is as shown in below: