- #1

Lambda96

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- Homework Statement
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- Relevant Equations
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Hi,

I am not sure if I have derived the matrix correctly, because of my results in task b

I solved task 1 as follows, I assumed that all three particles move to the right

$$m \dot{x_1}=-k(x_1 - x_2)$$

$$2m \dot{x_2}=-k(x_2-x_2)-3k(x_2-x_3)$$

$$3m \dot{x_3}=-3k(x_3-x_2)$$

Then I simply divided all three equations by the masses and got the following form

$$ \dot{x_1}=-\frac{k}{m}(x_1 - x_2)$$

$$ \dot{x_2}=-\frac{k}{2m}(x_2-x_2)-\frac{3k}{2m}(x_2-x_3)$$

$$ \dot{x_3}=-\frac{k}{m}(x_3-x_2)$$

Then I set up the 3 equations in the required matrix form:$$\frac{d}{dt^2} \vec{x}=\left( \begin{array}{rrr}

-\frac{k}{m} & \frac{k}{m} & 0 \\

\frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m}) & \frac{3k}{2m} \\

0 & \frac{k}{m} & -\frac{k}{m} \\

\end{array}\right) \left( \begin{array}{rrr}

x_1 \\

x_2 \\

x_3 \\

\end{array}\right)$$

For task part b, I simply put ##\vec{x}(t)=e^{i \omega t} \vec{v}## into the matrix above and then divided out the exponential term on both sides.$$ -\omega^2 \vec{v}= \left( \begin{array}{rrr}

-\frac{k}{m} & \frac{k}{m} & 0 \\

\frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m}) & \frac{3k}{2m} \\

0 & \frac{k}{m} & -\frac{k}{m} \\

\end{array}\right) \vec{v}$$

After that I just got everything on one site and used that ##\frac{k}{m}=\omega_0## is

$$ \left( \begin{array}{rrr}

0 \\

0 \\

0 \\

\end{array}\right)= \left( \begin{array}{rrr}

-\frac{k}{m}+\omega^2 & \frac{k}{m} & 0 \\

\frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m})+\omega^2 & \frac{3k}{2m} \\

0 & \frac{k}{m} & -\frac{k}{m}+\omega^2 \\

\end{array}\right) \vec{v}$$

Now, to determine ##\omega##, I simply formed the determinant of the matrix and got the following:

$$\frac{3 \omega_0^6}{2}+\frac{3 \omega_0^4 \omega^2}{2}-2 \omega_0^2 \omega^4- \omega^6=0$$

If I now solve the equation for ##\omega##, I get the following values

$$\omega_1=\pm \omega_0$$

$$\omega_2=\pm \sqrt{\frac{1}{2}(3+ \sqrt{3})}\sqrt{-\omega_0^2}$$

$$\omega_3=\pm \frac{\sqrt{(\sqrt{3}-3) \omega_0^2}}{\sqrt{2}}$$

I am not sure if I have derived the matrix correctly, because of my results in task b

I solved task 1 as follows, I assumed that all three particles move to the right

$$m \dot{x_1}=-k(x_1 - x_2)$$

$$2m \dot{x_2}=-k(x_2-x_2)-3k(x_2-x_3)$$

$$3m \dot{x_3}=-3k(x_3-x_2)$$

Then I simply divided all three equations by the masses and got the following form

$$ \dot{x_1}=-\frac{k}{m}(x_1 - x_2)$$

$$ \dot{x_2}=-\frac{k}{2m}(x_2-x_2)-\frac{3k}{2m}(x_2-x_3)$$

$$ \dot{x_3}=-\frac{k}{m}(x_3-x_2)$$

Then I set up the 3 equations in the required matrix form:$$\frac{d}{dt^2} \vec{x}=\left( \begin{array}{rrr}

-\frac{k}{m} & \frac{k}{m} & 0 \\

\frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m}) & \frac{3k}{2m} \\

0 & \frac{k}{m} & -\frac{k}{m} \\

\end{array}\right) \left( \begin{array}{rrr}

x_1 \\

x_2 \\

x_3 \\

\end{array}\right)$$

For task part b, I simply put ##\vec{x}(t)=e^{i \omega t} \vec{v}## into the matrix above and then divided out the exponential term on both sides.$$ -\omega^2 \vec{v}= \left( \begin{array}{rrr}

-\frac{k}{m} & \frac{k}{m} & 0 \\

\frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m}) & \frac{3k}{2m} \\

0 & \frac{k}{m} & -\frac{k}{m} \\

\end{array}\right) \vec{v}$$

After that I just got everything on one site and used that ##\frac{k}{m}=\omega_0## is

$$ \left( \begin{array}{rrr}

0 \\

0 \\

0 \\

\end{array}\right)= \left( \begin{array}{rrr}

-\frac{k}{m}+\omega^2 & \frac{k}{m} & 0 \\

\frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m})+\omega^2 & \frac{3k}{2m} \\

0 & \frac{k}{m} & -\frac{k}{m}+\omega^2 \\

\end{array}\right) \vec{v}$$

Now, to determine ##\omega##, I simply formed the determinant of the matrix and got the following:

$$\frac{3 \omega_0^6}{2}+\frac{3 \omega_0^4 \omega^2}{2}-2 \omega_0^2 \omega^4- \omega^6=0$$

If I now solve the equation for ##\omega##, I get the following values

$$\omega_1=\pm \omega_0$$

$$\omega_2=\pm \sqrt{\frac{1}{2}(3+ \sqrt{3})}\sqrt{-\omega_0^2}$$

$$\omega_3=\pm \frac{\sqrt{(\sqrt{3}-3) \omega_0^2}}{\sqrt{2}}$$

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