# Particular Solution of A Coupled and Driven Oscillator

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1. Oct 24, 2016

### PatsyTy

1. The problem statement, all variables and given/known data

Consider two masses m connected to each other and two walls by three springs with spring constant k. The left mass is subject to a driving force $F_d\cos(2 \omega t)$ and the right to $2F_d\cos(2 \omega t)$

2. Relevant equations

Writing out the coupled equations:

$$m_1 x_1''+2kx_1-kx_1 = F_d \cos (2\omega t)$$
$$m_2 x_2''-kx_1+2kx_2 = 2F_d\cos (2 \omega t)$$

3. The attempt at a solution

Assume a solution
$$x_1 = A_1 \cos (\omega t) \rightarrow x_1'' = -A_1 \omega^2 \cos (\omega t)$$
$$x_2 = A_2 \cos (\omega t) \rightarrow x_2'' = -A_2 \omega^2 \cos (\omega t)$$

Sub this into our original equation and write it as a matrix equation

$$-\omega^2 \cos (\omega t) \left( \begin{array}{c} A_1 \\ A_2 \end{array} \right) + \omega^2 \left( \begin{array}{cc} 2 & -1 \\ -1 & 2 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right) = \frac{F_d}{m} \cos (2 \omega t) \left( \begin{array}{c} 1 \\ 2 \end{array} \right)$$

where $\omega^2 = k/m$. This is where I get stuck, before we would form an eignevalue problem and solve the characteristic polynomial to get the eigenvalues then the constants $A_1$ and $A_2$ however we have too many terms to do this.

A suggestion on where to go from here would be greatly appreciated!

Last edited: Oct 24, 2016
2. Oct 24, 2016

### TSny

The frequency ω in the driving forces is not necessarily equal to $\sqrt{k/m}$. You should use another notation, such as $\omega_0^2 = k/m$.

Since the driving forces have frequency $2 \omega$, you might be better off with a trial solution where $x_1$ and $x_2$ vary with frequency $2 \omega$.

3. Oct 24, 2016

### PatsyTy

Thanks for the reply! That was my bad actually in copying the question, it does specify in the problem that the frequency of the driving for is equal to $\sqrt{k/m}$. I believe I managed to get the solution after a couple hours of work by adding the two equations and subtracting the two equations and making a substitution of variables:

$$(x_1''+x_2'')+\omega^2(x_1+x_2)=\frac{3F_d}{m} \cos (2 \omega t)$$
Set $z=x_1+x_2 \rightarrow z''=x_1''+x_2''$
Gives
$$z''+\omega^2 z = \frac{3F_d}{m} \cos (2 \omega t)$$

Similar argument for subtracting the two. Is it possible to solve this question using a matrix equation? I don't see any simple way of doing so but may not be seeing it.

Otherwise thanks for the help!

4. Oct 25, 2016

### TSny

Yes. Proceed as in your first post, but assume $x_1$ and $x_2$ vary as $\cos2\omega t$.