Answer:
[tex]y(t)=-\frac{25}{3}e^{-8t}+\frac{25}{3}e^{-2t}[/tex]
Step-by-step explanation:
We are given that
Initial velocity [tex]=v(0)=50m/s[/tex]
M=40 kg
K=64 N/m
C=40 Kg/s
We have to find the y(t)
Displacement equation for damped motion
[tex]M\frac{d^2y}{dt^2}+C\frac{dy}{dt}+ky=0[/tex]
Substitute the values
[tex]4\frac{d^2y}{dt^2}+40\frac{dy}{dt}+64y=0[/tex]
Auxillary equation
[tex]4m^2+40m+64=0[/tex]
[tex]m^2+10m+16=0[/tex]
[tex]m^2+8m+2m+16=0[/tex]
[tex]m(m+8)+2(m+8)=0[/tex]
[tex](m+8)(m+2)=0[/tex]
[tex]m+8\implies m=m_1=-8[/tex]
[tex]m+2=0[/tex]
[tex]\implies m=m_2=-2[/tex]
When roots are distinct and real
Then, [tex]y(t)=C_1e^{m_1t}+C_2e^{m_2t}[/tex]
Substitute the values
[tex]y(t)=C_1e^{-8t}+C_2e^{-2t}[/tex]
Substitute t=0
Then,[tex]y(0)=C_1+C_2[/tex]
Substitute y(0)=0 because when motion does not start then displacement y(t)=0
[tex]C_1=-C_2[/tex]
Differentiate w.r.t t
[tex]v(t)=y'(t)=-8C_1e^{-8t}-2C_2e^{-2t}[/tex]
Substitute the values v(0)=50 and t=0
[tex]50=-8C_1-2C_2[/tex]
Substitute the value
[tex]50=-8(-C_2)-2C_2=8C_2-2C_2=6C_2[/tex]
[tex]C_2=\frac{50}{6}=\frac{25}{3}[/tex]
[tex]C_1=-C_2=-\frac{25}{3}[/tex]
Substitute the values
[tex]y(t)=-\frac{25}{3}e^{-8t}+\frac{25}{3}e^{-2t}[/tex]
