Solution :
Given :
m = 2 kg
k = 8 N/m
C= 1 N-s/m
F(t) = 20 sin (4t)
[tex]$F_0= 20 \ N$[/tex]
ω = 4 rad/s
[tex]$\omega_n = \sqrt{\frac{k}{m}}$[/tex]
[tex]$\omega_n = \sqrt{\frac{8}{2}}$[/tex]
= 2 rad/s
Therefore,
[tex]$\epsilon = \frac{C}{2 m \omega_n}$[/tex]
[tex]$\epsilon = \frac{1}{2 \times 2 \times 2}$[/tex]
= [tex]$\frac{1}{8}$[/tex]
= 0.125
So, r = [tex]$\frac{\omega}{\omega_n}$[/tex]
= [tex]$\frac{4}{2}$[/tex]
= 2
Steady state response is given by
[tex]$(A)=\frac{F_0 / k}{\sqrt{(1-r^r)^2+(2 \epsilon r)^r}}$[/tex]
[tex]$(A)=\frac{20 / 8}{\sqrt{(1-2^r)^2+(2 \times 0.125 \times 2)^r}}$[/tex]
A = 0.82 m
