Answer:
Step-by-step explanation:
Given the mass is m =16kg, and 1N force will stretch the spring 1 m.
That is, F =1N,Z =1m. Now find the spring constant k:
F = kL = 1 = k(1) = k= 1N/m.
The damping force is 8times the instantaneous velocity, this means β = 8,
and the external force is f(t) = 0
Initially the object compressed 0.6m above equilibrium position,
with the downward velocity is 2m/s.
The differential equation for a spring mass system with
damping force and extemal force is: mx" + βxt + kx = f(t).
so, 16x"+ 8x' + x= 0, x(0} = -0.6, x'(0)= 2m/s.
Now solve the DE:
The auxilary equation for the homogeneous equation is 16x"+8x'+x=0
solving we get, 16r² + 8r + 1 = 0 => (4r + 1)² = 0 => r = - 1/4.
Then the general solution for the homogenous system is: [tex]x(t)=c_1e^{-t/4} +c_2te^{-t\4}[/tex].
Use the initial conditions x (0) = -0.6, x'(0) = 2m/s:
[tex]x(0)=c_1e^{0} +c_2(0)e^{0}=-0.6=c_1\\x'(t)=-\frac{1}{4}c_1e^{-t/4}+c_2e^{-t/4}-\frac{1}{4}c_2te^{-t/4}\\x'(0)=-\frac{1}{4}c_1e^0+c_2e^0-\frac{1}{4}c_2(0)e^0=2=-\frac{1}{4}(-0.6)+c_2=c_2=1.85[/tex].
Hence, [tex]x(t) =-0.6e^{-t/4}+1.85te^{-t/4}[/tex].
