Answer
$MF=0.997$
Work Step by Step
The required magnification factor can be determined as follows:
$\omega_n=\frac{K}{n}$
$\implies \omega_n=\frac{75\times 32.2}{7}=18.5742rad/s$
Now $MF=\frac{1}{\sqrt{(1-(\frac{\omega_{\circ}}{\omega_n})^2)^2+(2(\frac{c}{c_{\circ}})(\frac{\omega_{\circ}}{\omega_n}))^2}}$
We plug in the known values to obtain:
$MF=\frac{1}{\sqrt{(1-(\frac{2}{18.5742})^2)^2+(2(0.8)(\frac{2}{18.5742}))^2}}$
This simplifies to:
$MF=0.997$
