Answer:
400 N
Explanation:
[tex]\text { Given: } m=300 \mathrm{~kg}, k=40,000 \mathrm{~N} / \mathrm{m}, \omega_{b}=\omega_{n}(r=1), X=10 \mathrm{~mm}, Y=2.5 \mathrm{~mm}[/tex] .
Find damping constant
[tex] \frac{X}{Y}=\left[\frac{1+(2 \zeta r)^{2}}{\left[\left(1-r^{2}\right)^{2}+(2 \zeta r)^{2}\right.}\right]^{1 / 2} \\ \left.\frac{10}{2.5}=\frac{\left\lceil 1+4\zeta^{2}\right]^{1 / 2}}{4 \zeta^{2}}\right] \\ 16=\frac{1+4 \zeta^{-2}}{4 \zeta^{2}} \\ \zeta^{2}=\frac{1}{60}=\frac{c^{2}}{4 k m} \\ c=\sqrt{\frac{4(40,000)(300)}{60}} \\ c=894.4 \mathrm{~kg} / \mathrm{s}[/tex]
Amplitude of force on base:
[tex]F_{T}=k Y r^{2}\left[\frac{1+(2 \zeta r)^{2}}{\left(1-r^{2}\right)^{2}+(2 \zeta r)^{2}}\right]^{1 / 2}[/tex]
substituting the values in above formula we get
F_T = 400 N
