## Essential University Physics: Volume 1 (3rd Edition)

We know the following equation: $A(\omega)=\frac{F_0}{m\sqrt{(\frac{b}{m})^2\omega_0^2+(w_0^2-w_d^2)^2}}$ Plugging in $1.1\omega_0$ into the equation and dividing by 5 (in the original equation) gives: $=\frac{3.29}{5}=\fbox{65.8 percent}$ Plugging in $.9\omega_0$ into the equation and dividing by 5 (in the original equation) gives: $=\frac{3.82}{5}=\fbox{76.4 percent}$