m = 4
k = 25.56
b = 20
[tex]x_o=1\\v_o=0[/tex]
The differential equation for damping motion is:
[tex]x''+\gamma x'+\omega_o^2 x=0[/tex]
where,
[tex]\gamma = \frac{b}{m} = \frac{20}{4}=5[/tex]
[tex]\omega_o^2=\frac{k}{m} = \frac{25.56}{4}=6.39[/tex]
Substitute the values in the differential equation and consider x" = r², x' =r and solve:
[tex]x''+5x'+6.39x=0\\r^2+5r+6.39=0\\r=-2.5 \pm0.37i[/tex]
Therefore, solution is given by:
[tex]x(t) = e^{-2.5t}[C_1cos0.37t+C_2sin0.37t][/tex][tex]\\[/tex]
at t = 0, x = 1
[tex]C_1=1[/tex]
[tex]x'(t) =-2.5e^{-2.5t}[C_1cos0.37t+C_2sin0.37t]+e^{2.5t}[-0.37C_1sin0.37t+0.37C_2cos0.37t][/tex]
at t = 0
x' =0
[tex]\\[/tex][tex]C_2=6.76[/tex]
[tex]x(t) = e^{-2.5t}[cos0.37t+6.76sin0.37t][/tex]
