# Two spring-coupled mass system

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 $$\omega = \sqrt{\frac{2k_0k}{m} + (\frac{k_0}{m})^2}$$

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.

• When trying to figure out the equations of motion, it's always best to move both masses in the same direction. The solutions work no matter where the masses are, since they only deal with relative displacements, but it's easier if all the quantities are positive. The mistake you're making is ignoring that in the first case, $x_1< 0$, since it's moved to the left, rather than the right. Otherwise, everything else is correct. Jan 28 at 11:34
• Please define the quantities $x_1$ and $x_2$, they're not defined in the picutre. Specifying that will also help you see yourself which distances are relevant for the springs. Can you tell us about the logic that brought you to either of the equations of motion? Then we can discuss where your logic is flawed and where it is correct, so that we can help you with the problem solving strategy. Jan 28 at 11:41
• Also, you have a $k$ missing in your first set of equations, and I think the second set of equations you meant to write "$-kx_1$" instead of "$+kx_1$". Jan 28 at 11:45
• Yeah drawing a free body diagram where not all positive displacements are on the same direction is a major faux-pas for this reason exactly. The same can be said about angles too. I suggest you do it the conventional way and only where you input initial conditions you consider the displacement of the first mass a negative value. Jan 29 at 19:06
• If $x_1$ is acting on a negative sense, then check that $m \ddot{x}_1$ is correct. You have to use $-m \ddot{x}_1$ I think since forces are still (+) to the right. Jan 29 at 19:12

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.

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 $$\frac{\textrm{d}^2}{\textrm{d}t^2}\begin{pmatrix} x_1' \\ x_2'\end{pmatrix} = \begin{pmatrix} - (k + k_0) & k \\ k & -(k + k_0)\end{pmatrix}\begin{pmatrix} x_1' \\ x_2'\end{pmatrix}.$$

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

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

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.