Two spring-coupled mass system

Question

I am confused with writing the equation of motion of masses and finding normal modes.

The problems I've dealing with before, masses is always moved in same directions, and I determined if springs are stretched or compressed using $(x_2-x_1)$.

But in this, the mass $m_1$ moved to the left and the mass $m_2$ moved to the right. So springs (with constant $k_0$) are compressed(springs push masses to their equilibrium), then spring (with $k$ in mid is stretched (pulls the masses to each other).

For $m_1 = m_2$ and $x_1 \ne x_2 $ If I write and solve that EoMs $$ m\ddot{x}_1 = +k_0x_1+k(x_1 + x_2)\\ m\ddot{x}_2 = -k_0x_2-k(x_1+x_2) $$ I get $$ \omega = \sqrt[4]{\frac{2k_0k}{m} + (\frac{k_0}{m})^2}$$

In addition, I write another pair of EoM: $$ m\ddot{x}_1 = +k_0x_1+k(x_2 - x_1)\\ m\ddot{x}_2 = -k_0x_2 -k(x_2 - x_1) $$

then gives

$$ \omega_1 = \sqrt{\frac{k_0-k}{m}} \\ \omega_2 = \sqrt{\frac{k_0+k}{m}} $$ I am not sure which way is correct? I am really obsessed with that and want to figure it out. I need your help :) Thanks in advance.

Answer 1

Thanks for defining $x$, with that, I can explain what the problem is. I assume that by the way you drew $x_1$ and $x_2$, this also defines the positive values of $x_1$ and $x_2$.

The thing is that you give $x_1$ and $x_2$ with reference to some equilibrium positions $x_{1e}$ and $x_{2e}$. You can define the harmonic motion in this coordinate system, but it's a bit tricky to justify exactly why the terms aris the way they do.

The no-brain way of doing it is to start with $x_1$ and $x_2$ with reference to the same origin. Say $x_1$ were the distance of mass 1 from the point where the left spring is attached to the wall, and $x_2$ were the distance of mass 2 from the same point. Then you can go ahead and model all three springs. The length of the first spring would be $x_1$, the length of the second spring would be $x_2 - x_1$ and the lenght of the third spring is $L - x_2$. Clearly, $x_1$ feels the forces due to the first and second spring, and each force is the spring constant times the length, so $$m \ddot{x}_1 = - k_0 x_1 + k (x_2 - x_1).$$ The signs are there because the left spring pulls $x_1$ to the left (against the positive direction of the coordinate system we just chose) while the middle spring gives rise to a pull to the right (in the positive direction of the coordinate system). Similarly, you find $$m \ddot{x}_2 = k_0 (L - x_2) - k (x_2 - x_1).$$ This time, the right spring pulls to the right while the middle spring pulls to the left, so you get the opposite signs. Therefore, you have a matrix equation $$\frac{\textrm{d}^2}{\textrm{d}t^2}\begin{pmatrix} x_1 \\ x_2\end{pmatrix} = \begin{pmatrix} - (k + k_0) & k \\ k & -(k + k_0)\end{pmatrix}\begin{pmatrix} x_1 \\ x_2\end{pmatrix} + k_0 L \begin{pmatrix} 0 \\ 1\end{pmatrix},$$ and you can proceed to calculate the frequencies from the frequency matrix.

Link between descriptions

I had kind of a brainblockage yesterday, so I got a bit confused about the use of different equilibrium positions. So to get this straight: You can transform the above equation to arbitrary coordinate origins for $x_1$ and $x_2$. Let's write new coordinates with the transform $x_1 \mapsto x_1' = x_1 - x_{1e}$ and $x_2 \mapsto x_2' = x_2 - x_{2e}$. Evidently, $\textrm{d}^2/\textrm{d}t^2 x_i' = \textrm{d}^2/\textrm{d}t^2 x_i$ for $i \in \{1,2\}$, and by insertion we have $$\frac{\textrm{d}^2}{\textrm{d}t^2}\begin{pmatrix} x_1' \\ x_2'\end{pmatrix} = \begin{pmatrix} - (k + k_0) & k \\ k & -(k + k_0)\end{pmatrix}\begin{pmatrix} x_1' + x_{1e} \\ x_2' + x_{2e}\end{pmatrix} + k_0 L \begin{pmatrix} 0 \\ 1\end{pmatrix}.$$ Now if you deliberately choose the positions $x_{1e}$ and $x_{2e}$ to be the equilibrium posistion, i.e. $$\begin{pmatrix} - (k + k_0) & k \\ k & -(k + k_0)\end{pmatrix}\begin{pmatrix} x_{1e} \\ x_{2e}\end{pmatrix} + k_0 L \begin{pmatrix} 0 \\ 1\end{pmatrix} = 0,$$ which is just the requirement that the masses feel no acceleration in this particular configuration, you arrive at $$\frac{\textrm{d}^2}{\textrm{d}t^2}\begin{pmatrix} x_1' \\ x_2'\end{pmatrix} = \begin{pmatrix} - (k + k_0) & k \\ k & -(k + k_0)\end{pmatrix}\begin{pmatrix} x_1' \\ x_2'\end{pmatrix}.$$

Answer 2

If you just have one spring with mass you have two possibilities to choose the force sign $~+~$ or $~-~$

$$m\ddot{x}=\pm k\,x$$

if you choose $~-~$ sign you obtain the solution with the initial conditions $x(0)=x_0$ and $\dot{x}(0)=0$

$$x(t)=x_0\,\cos(\omega\,t)$$

where $\omega=\sqrt{\frac km}$

if you choose $~+~$ sign you obtain the solution

$$x(t)=\frac 12 x_0 e^{\omega t}$$

but this solution is wrong because you expect sine or cosine motion.

thus the minus sign is correct.

follow this rule you obtain this equations of motion

$$m\,\ddot x_1=-k_0\,x_1-k\,(x_1-x_2)$$ $$m\,\ddot x_2=-k_0\,x_2+k\,(x_1-x_2)$$

Answer 3

I suggest stick to convention and do the free body diagram with positive displacements towards the right.

In fact I drew the above with $x_2 > x_1$ to help with the determination of the middle spring force direction.

From the above I have

$$ \begin{aligned} F_1 & = k_0 (x_1) \\ F_2 & = k (x_2-x_1) \\ F_3 & = k_0 (x_2) \end{aligned} $$

and the equations of motion

$$ \begin{aligned} F_2 - F_1 & = m_1 \ddot{x}_1 \\ -F_2 - F_3 & = m_2 \ddot{x}_2 \end{aligned} $$

Combined the above produces

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_1 \end{bmatrix} \pmatrix{\ddot{x}_1 \\ \ddot{x}_2 } = -\begin{bmatrix} k+k_0 & -k \\ -k & k+k_0 \end{bmatrix} \pmatrix{x_1 \\ x_2} $$

Now you can go and flip the sign of $x_1$ and of $\ddot{x}_1$ if you want.