Answer: [tex]x(t) = 14e^{-2t} - 10e^{-3t}[/tex]
Explanation: In a mass-spring-damper system, the differential equation that rules the motion of the mass is: mx" + cx' + kx = 0
Using m = 4, k = 24 and c = 20, we have
4x" + 20x' + 24x = 0
Simplifying, we have
x" + 5x'+ 6x = 0
The characteristic equation of this differential is
[tex]r^{2} + 5r + 6 = 0[/tex]
The solutions for the quadratic equation are: [tex]r_{1} = -2[/tex] and [tex]r_{2} = -3[/tex]
Hence:
x(t) = [tex]C_{1}e^{-2t} + C_{2}e^{-3t}[/tex]
x'(t) = [tex]-2C_{1}E^{-2t} - 3C_{2}e^{-3t}[/tex]
To determine the constants, we have the initial conditions x(0) = 4 and
x'(0) = 2, then:
[tex]x(0) = C_{1} + C_{2} = 4\\ C_{1} = 4 - C_{2}[/tex]
[tex]x'(0) = -2C_{1} -3C_{2} = 2\\-2(4-C_{2}) -3C_{2} = 2\\C_{2} = -10\\C_{1} = 4 - C_{2}\\C_{1} = 14[/tex]
Substituing the constants:
[tex]x(t) = 14e^{-2t} - 10e^{-3t}[/tex]
The position function for this system is: [tex]x(t) = 14e^{-2t} - 10e^{-3t}[/tex]