Finding Steady State Solutions for a Mass-Spring System with a Moving Support

[Benny]

Can someone help me out with the following question?

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

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

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 x_p \left( t \right) = \frac{1}{5}\left( {4\sin t - \cos t} \right).

b) I'm not really sure about this part but I found a particular solution anyway. I obtained x_p \left( t \right) = \frac{{2\sin \left( t \right)}}{{\omega _0 ^2 - 1}}. 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 dL = \sqrt {1 + \left( {\frac{{dy}}{{dx}}} \right)^2 } dx. A 'sample' circumference is dC = 2\pi (height) = 2\pi f\left( x \right). Then the surface area of the graph revolved about the x-axis from x = a to x = b is S = \int\limits_a^b {2\pi f\left( x \right)} \sqrt {1 + \left( {\frac{{dy}}{{dx}}} \right)^2 } dx?

 

[saltydog]

Well Benny, this involves resonance right? What happens when \omega=1 in the differential equation with no damping? We get:

x^{&#039;&#039;}+x=2Sin(t)

But in that case, both the roots of the auxilliary equation and the roots of the differential operator for which the RHS is a particular solution are \pm i? That means we have double roots. That means the particular solution is actually of the form:

y_p(x)=AtSin(t)+BtCos(t)

So, the t in front of them means it oscillates without bounds right? That happens whenever input frequenies match output frequencies.

 

[Benny]

Thanks for the help Saltydog.