Answer:
[tex]u(t)=\frac{1}{12\sqrt{31}}e^{-2t}sin(2\sqrt{31}t)[/tex]
Explanation:
Our values given are,
[tex]L=3in=0.25ft[/tex]
Ramping constant
[tex]r=2lb.sec/ft[/tex]
By this way we can calculate k and m.
[tex]m=\frac{mg}{g} = \frac{16}{32}=\frac{1}{2} lb/ft/sec^2[/tex]
[tex]k=\frac{mg}{L} = \frac{16}{1/4} = 64lb/ft[/tex]
We can now create the differential equation, which is given by,
[tex]mu''+ru'+ku=0[/tex]
[tex]\Rightarrow \frac{1}{2}u''+2u'+54u=0[/tex]
The initial conditions are:
[tex]u(0) = 0[/tex]
[tex]u'(0) = \frac{1}{6}[/tex]
*Note that the velocity war given in inches and transformated to feet.
With that we can create our characteristic equation:
[tex]\frac{1}{2}r^2+2r+64=0[/tex]
With two roots
[tex]r_1 = -2+2\sqrt{31}i[/tex]
[tex]r_2 = -2-2\sqrt{31}i[/tex]
Our general solution of the Differential equation is then,
[tex]u(t) = e^{2t}[c_1 cos(2\sqrt{31}t)+c_2 sin(2\sqrt{31}t)][/tex]
Applying the initial conditions,
[tex]u(t)=0\Rightarrow 0=c_1[/tex]
[tex]u'(t)=\frac{1}{6} \Rightarrow \frac{1}{6} = e^{0}(2\sqrt{31}c_2)[/tex]
The position of u at any time t is,
[tex]u(t)=\frac{1}{12\sqrt{31}}e^{-2t}sin(2\sqrt{31}t)[/tex]
