Respuesta :
Answer:
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
[tex]my'' + \zeta y' + ky=0[/tex]
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
[tex]w - kL=0[/tex]
[tex]mg - kL=0[/tex]
[tex]4 - k\frac{1}{6}=0[/tex]
[tex]k = 4\times 6 =24[/tex]
The damping constant can be found by
[tex]F = -\zeta y'[/tex]
[tex]6 = 3\zeta[/tex]
[tex]\zeta = \frac{6}{3} = 2[/tex]
Finally, the mass m can be found by
[tex]w = 4[/tex]
[tex]mg=4[/tex]
[tex]m = \frac{4}{g}[/tex]
Where g is approximately 32 ft/s²
[tex]m = \frac{4}{32} = \frac{1}{8}[/tex]
Therefore, the required differential equation is
[tex]my'' + \zeta y' + ky=0[/tex]
[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]
The initial position is
[tex]y(0) = \frac{1}{2}[/tex]
The initial velocity is
[tex]y'(0) = 0[/tex]
