# Can we let the lowest of n by equal (lenght and k) springs connected masses in equilibrium move in a siusoid way after giving the lowest a pull?

Imagine we hang n masses, connected by equal springs of equal length and with equal k (suspended on a very high ceiling or whatever what, as long as it doesn't exchange energy with the system). So what we see are different masses hanging vertically and connected by equal length pieces of springs, all with the same k. So if a heavy mass hangs under two lighter masses the spring on which it's attached is stretched more.

When the system is in equilibrium (on earth) and we pull the lowest mass down after which releasing it, can we let the bottom mass move in a sinusoid way by adjusting the masses in the right way, or will the system always show an irregular, or periodic (but non-sinusoid) motion?

I know the system is highly non-linear, and thus difficult to solve mathematically, but can't we nevertheless reason by intuition if it's possible or not.

I can't imagine that there is not one manner to adjust those masses so the whole chain will make a sinusoid motion, by which I mean the motion the lowest mass makes.

• AFAIK if they are ideal springs, the system will have a periodic oscillation in the total centre of mass. I'm not sure how you would mathematically prove the individual masses are periodic or non-periodic though. It would depend on factors like the number of masses and such.
– JMac
May 19 '17 at 14:24
• For small oscillations, one can obtain the normal modes of the system, in which each degree of freedom is executing sinusoidal motion. This is independent of the masses or any other specifics if the system. However, these degrees of freedom will in general not be the observed displacements, but rather some linear combinations. If you wish to know when the normal mode matrix will be diagonal, i e each degree of freedom is independently oscillating in a sinusoid, it would happen only when they decouple. i. e. if the blocks are coupled, there is no way they would execute sinusoidal motion. May 19 '17 at 14:30
• That the blocks will make a sinusoid motion if decoupled is, of course, true, but why shouldn't there exist, by the right (out of an infinite number) choice of the block's masses, a motion for the lowest block that is sinusoid? May 19 '17 at 15:15
• To be clear, are the rest of the masses free to move? Or do you want them to remain static? May 23 '17 at 8:13
• Yes, all the masses are free to move after the whole chain (of which the masses are separated at an equal distance by springs with the same k) is pulled down a certain distance after which it is released. At equilibrium, the springs of equal length are all under different tension. First I thought that all the masses would make a sinusoid movement in this case, but the eigentimes of the oscillations of separate masses vary with $\sqrt m$, where m is, of course, the mass of one of the masses. May 23 '17 at 11:41

The answer is to the negative according to the following counterexample for $n=2\;$ :
If $x_1\left(t\right),x_2\left(t\right)$ are the vertical displacements of the top and bottom masses $m_1,m_2$ from their equilibrium positions respectively (and not from the ceiling), the equations of motion are \begin{align} m_1\ddot{x}_1 & =+k\left(x_2-2x_1\right) \tag{01.1}\\ m_2\ddot{x}_2 & =-k\left(x_2-x_1\right) \tag{01.2} \end{align} and by matrices $$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{bmatrix} + k\begin{bmatrix} +2 & -1 \\ -1 & +1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \tag{02}$$ that is $$\mathrm{M}\ddot{\mathbf{x}}+\mathrm{K}\mathbf{x}=\mathbf{0} \tag{03}$$ where $$\mathbf{x} \equiv \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} , \quad \mathrm{M} \equiv \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} , \quad \mathrm{K} \equiv k\begin{bmatrix} +2 & -1 \\ -1 & +1 \end{bmatrix} \tag{04}$$ Since $\mathrm{M}$ is invertible $(m_1m_2 \ne 0)$, multiplying equation (03) by $\mathrm{M}^{-1}$ we have $$\ddot{\mathbf{x}}+\left(\mathrm{M}^{-1}\mathrm{K}\right)\mathbf{x}=\ddot{\mathbf{x}}+\mathrm{S}\mathbf{x}=\mathbf{0} \tag{05}$$ where $$\mathrm{S}\equiv \mathrm{M}^{-1}\mathrm{K}= \begin{bmatrix} m_1^{-1} & 0 \\ 0 & m_2^{-1} \end{bmatrix} k \begin{bmatrix} +2 & -1 \\ -1 & +1 \end{bmatrix} =\dfrac{k}{m_2} \begin{bmatrix} 2\dfrac{m_2}{m_1} & -\dfrac{m_2}{m_1} \\ -1 & +1 \end{bmatrix} \tag{06}$$ or $$\mathrm{S} =\omega_{0}^2 \begin{bmatrix} 2\mu & -\mu\\ -1 & +1 \end{bmatrix} \;, \quad \omega_{0}\equiv\sqrt{\dfrac{k}{m_2}}\; , \quad \mu \equiv \dfrac{m_2}{m_1} =\text{mass ratio} \tag{07}$$ For the eigenvalues $\xi$ of the matrix $$\dfrac{\mathrm{S}}{\omega_{0}^2}= \begin{bmatrix} 2\mu & -\mu\\ -1 & +1 \end{bmatrix} \tag{08}$$ we have $$\begin{vmatrix} 2\mu-\xi & -\mu\\ -1 & 1-\xi \end{vmatrix} =0 \quad \Longrightarrow \quad \xi^2-\left(2\mu+1\right)\xi +\mu=0 \tag{09}$$ So, the eigenvalues are the following positive real numbers \begin{align} \xi_1 & =\dfrac{\left(2\mu+1\right)+\sqrt{4\mu^2+1}}{2}\in \mathbb{R}^{+} \tag{10.1}\\ \xi_2 & =\dfrac{\left(2\mu+1\right)-\sqrt{4\mu^2+1}}{2}\in \mathbb{R}^{+} \tag{10.2} \end{align} Now if $$\mathrm{P}= \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix} \tag{11}$$ is an invertible real matrix which diagonalizes the matrix of equation (08), then $$\mathrm{P}^{-1} \begin{bmatrix} 2\mu & -\mu\\ -1 & +1 \end{bmatrix} \mathrm{P} = \begin{bmatrix} \xi_1 & 0 \\ 0 & \xi_2 \end{bmatrix} \tag{12}$$ and consequently $$\mathrm{P}^{-1}\mathrm{S}\mathrm{P} = \omega_{0}^2 \begin{bmatrix} \xi_1 & 0 \\ 0 & \xi_2 \end{bmatrix} \tag{13}$$ Defining $$\mathbf{y}\equiv\mathrm{P}^{-1}\mathbf{x} \tag{14}$$ equation (05) gives after multiplication by $\mathrm{P}^{-1}$ $$\ddot{\mathbf{y}}+\left(\mathrm{P}^{-1}\mathrm{S} \mathrm{P}\right)\mathbf{y}=\mathbf{0} \tag{15}$$ or $$\begin{bmatrix} \ddot{y}_1 \\ \ddot{y}_2 \end{bmatrix} + \omega_{0}^2 \begin{bmatrix} \xi_1 & 0 \\ 0 & \xi_2 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \tag{16}$$ that is 2 independent differential equations \begin{align} \ddot{y}_1+\xi_1\omega_{0}^{2} y_{1} & =0 \tag{17.1}\\ \ddot{y}_2+\xi_2\omega_{0}^{2} y_{2} & =0 \tag{17.2} \end{align} with general solutions respectively \begin{align} y_{1}\left(t\right) & = A_{1}\sin\left(\omega_1 t+\phi_{1}\right) \tag{18.1}\\ y_{2}\left(t\right) & = A_{2}\sin\left(\omega_2 t+\phi_{2}\right) \tag{18.2} \end{align} where \begin{align} \omega_1 & =\sqrt{\xi_1}\,\omega_0 =\sqrt{\dfrac{\left(2\mu+1\right)+\sqrt{4\mu^2+1}}{2}}\,\omega_0 \tag{19.1}\\ \omega_2 & =\sqrt{\xi_2}\,\omega_0 =\sqrt{\dfrac{\left(2\mu+1\right)-\sqrt{4\mu^2+1}}{2}}\,\omega_0 \tag{19.2} \end{align} The motion of the system is the composition of two independent harmonic(sinusoid) oscillations of angular frequencies $\omega_1,\omega_2$ \begin{align} x_{1}\left(t\right)& = p_{11}\cdot y_{1}\left(t\right)+ p_{12}\cdot y_{2}\left(t\right)=p_{11} A_{1}\sin\left(\omega_1 t+\phi_{1}\right)+p_{12} A_{2}\sin\left(\omega_2 t+\phi_{2}\right) \tag{20.1}\\ x_{2}\left(t\right)& = p_{21}\cdot y_{1}\left(t\right)+ p_{22}\cdot y_{2}\left(t\right)=p_{21} A_{1}\sin\left(\omega_1 t+\phi_{1}\right)+p_{22} A_{2}\sin\left(\omega_2 t+\phi_{2}\right) \tag{20.2} \end{align} The constants $A_1,\phi_1,A_2,\phi_2$ are determined from the initials conditions, for example from the positions and velocities of the two masses at time $t=0$ : $x_1\left(0\right),\dot{x}_1\left(0\right),x_2\left(0\right),\dot{x}_2\left(0\right)$.
• @descheleschilder Of course the case $n=2$ herein is a first indication not a proof for the impossibility of sinusoid motion. Intuitively I think that for $n>2$ the impossibility is even worse. Also, I think that a group of masses doesn't give combined motion to an other group. The motion of any mass is a combination (superposition) of independent harmonic oscillations (the so-called normal modes). By the way, if a "nice answer" of mine doesn't answer your question, please don't upvote it, since moreover I don't care so much about reputation. May 23 '17 at 12:42