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I am stuck part way through the following and not sure how or if finding eigenvalues will help with finding modes of oscillations:

Consider the system of three masses and two ideal elastic bands:

$(m)$---$k$---$(2m)$---$2k$---$(m)$ [$m$ are masses, $k$ is spring constant]

Find the resonance frequencies of oscillation, normal modes of oscillation and describe motion of masses corresponding to modes of oscillations.

My work so far:

$$V=\frac{1}{2}(k9x-y)^{2}+k(y-z)^{2}=\frac{1}{2}k[x^{2}-2xy+3y^{2}-4yz+2z^{2}]$$ $$m\ddot{x}=-\frac{\partial V}{\partial x}=-k(x-y)$$ $$m\ddot{y}=-\frac{\partial V}{\partial y}=-k(-x+3y-2z)$$ $$m\ddot{z}=-\frac{\partial V}{\partial z}=-k(2z-2y)$$ $$\therefore\ \ m\ddot{x}+m\ddot{y}+m\ddot{z}=0$$ $$y=\frac{-1}{2}(x+z)$$ $$\therefore -m\omega^{2}x=-\frac{1}{2}k(3x+z)$$ $$\therefore -m\omega^{2}y=-k(x+3z)$$ Up to here I'm good. But from here, to find the normal modes of oscillation I am not sure where to go. If I define $\lambda=\frac{m\omega^{2}}{k}$(from an example in text) I get: $$\lambda(X) = \left[ \begin{array}{cc} \frac{3}{2} & \frac{1}{2} \\ 1 & 3 \\ \end{array} \right](X)$$ Would I then use $(A-\lambda I)(X)=0$ to find eigenvalues? If I do so, I get: $\lambda_{1,2}=\frac{1}{4}(9 \pm \sqrt{17})$. At this point I begin to lose confidence and understanding in my method. Any help or guidance would be appreciated. Also, I apologize if the formatting is off putting, I am still learning the language.

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Without using the condition $m\ddot{x}+m\ddot{y}+m\ddot{z}=0$, just write your 3D system as $$ k\left[ \begin {array}{ccc} -1&1&0\\1&-3&2 \\ 0&2&-2\end {array} \right] \left[ \begin {array}{c} x\\y\\z\end {array} \right]=m\left[ \begin {array}{c} \ddot{x}\\\ddot{y}\\\ddot{z}\end {array} \right]$$

Now compute the frequencies as eigenvalues and normal modes of oscillation as eigenvectors.

Notice that the eigenvalue $0$, corresponding to uniform motion of the system as a whole, appears naturally.

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  • $\begingroup$ I guess I've not come across (that I can remember) an eigenvalue problem like this (vector $x$ on one side and vector $\ddot{x}$ on the other). Would it still be of the form $(\frac{k}{m}A-\lambda I)(X)=0$ where (X) is {x,y,z}? $\endgroup$
    – NotSoSN
    Mar 12, 2016 at 1:32
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    $\begingroup$ We start with 3 equations which in matrix notation are $M\vec{v}=\ddot{\vec{v}}$. We want to change basis in order to decouple these equations and arrive at 3 equations of the form $\lambda_i X_i=\ddot{X_i}$. This decoupling is what diagonalization of $M$ provides. $\endgroup$
    – Marcel
    Mar 12, 2016 at 1:36

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