Answer
10 percent more: 65.8 percent
10 percent less: 76.4 percent
Work Step by Step
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:
$=\frac{3.82}{5}=\fbox{76.4 percent}$
