Q. The position x(t) at time t of a mass attached to a spring hanging from a moving support satisifies the differential equation:

[tex]

\frac{{d^2 x}}{{dt^2 }} + 2p\frac{{dx}}{{dt}} + \omega _0 ^2 x = 2\sin \left( t \right)

[/tex]

a) Find the steady state solution when w_0 = 3 and p = 1.

b) If p = 0 then there is a value for w_0 > 0 for which there is no steady state. What is this value of w_0? Justify your answer by finding the particular solution.

a) The auxillary equation has complex roots with a negative real part so the complimentary function isn't a part of the steady state solution since the decaying exponential leads to the complimentary function tending to zero as t gets large? So I need a particular solution I think. I found [tex]x_p \left( t \right) = \frac{1}{5}\left( {4\sin t - \cos t} \right)[/tex].

b) I'm not really sure about this part but I found a particular solution anyway. I obtained [tex]x_p \left( t \right) = \frac{{2\sin \left( t \right)}}{{\omega _0 ^2 - 1}}[/tex]. I don't understand what is meant by find a value for which there is no steady state. w_0 is constant so x_p(t) is just sine function with the 'usual' behaviour isn't it? If I were to guess I'd just say w_0 = 1 but could someone help me out with this question?

I would also like to know if the following would be a 'valid' way to quickly formulate the formula for the surface area of a graph revolved about the x-axis.

A bit of arc length is [tex]dL = \sqrt {1 + \left( {\frac{{dy}}{{dx}}} \right)^2 } dx[/tex]. A 'sample' circumference is [tex]dC = 2\pi (height) = 2\pi f\left( x \right)[/tex]. Then the surface area of the graph revolved about the x-axis from x = a to x = b is [tex]S = \int\limits_a^b {2\pi f\left( x \right)} \sqrt {1 + \left( {\frac{{dy}}{{dx}}} \right)^2 } dx[/tex]?