Slinky

Check out this video:

and two more after it for the answer.

Seems like a nice textbook problem to establish a simple model for the slinky. This is one I came up with: Suppose the slinky consists of \( N + 1 \) identical masses of mass \( m \) connected to their neighbor by a spring of spring constant \( k \). Suppose the position of mass \( j \) is \( x_j(t) \) and \( x_0(0) = 0 \). For convenience, let \( \Omega^2 = \frac{k}{m} \).

Before the drop

What is the equilibrium positions of the masses before the drop. It is actually quite simple from intuition. The spring above the j-th mass must hold up the weight of the \( N + 1 -j \) masses below it. So \( x_{j} - x_{j-1} = (N + 1 - j)\frac{g}{\Omega^2} \). This difference equation solves to:

\[ x_{j} (0) =\frac{g}{\Omega^2} \frac{(2N-j+1)j}{2} \]

So the equilibrium length of the slinky is: \( x_{N}(0) = \frac{g}{\Omega^2} \frac{N(N+1)}{2} \). From now on we'll work in unit where \( x_{N}(0) = g = 1\). This is a natural unit system for this problem. By the way, this suggests a characteristic collapse time of the slinky: \( \sqrt{x_{N}(0) / g} \). So if I estimate the equilibrium length of the slinky to be about 1 meter, then the collapse time is roughtly 0.32 seconds, which is reasonable judging from the video. In this unit,

\[ \Omega^2 = \frac{N(N+1)}{2} \]

The center of mass of the slinky is at

\[ x_{\rm cm}(0) = \frac{2}{N(N+1)} \cdot \frac{(N+1)(2N+1)}{6} = \frac{2N+1}{3(N+1)} \,. \]

Note that \( \lim_{N \to \infty} x_{\rm cm}(0)= \frac{2}{3} \), so that in the continuum limit the center of mass is at \( 1/3 \) the length of the slinky from the bottom. From this you can already partial see the correct answer: most of the mass is concentrated at the bottom so as the top is release, it is a lot easier to move the mass on the top portion than the bottom portion. Also the net force on every particle is \( 0 \) at \( t = 0 \). So immediately after the release, only the top mass, with the force holding it up vanished, has any appreciable acceleration. All the other masses still think they are still instantaneously in static equilibrium.

Equation of motion

Coming straight from the Lagrangian:

\[ L = \frac{m}{2} \sum_{j=0}^{N} \dot{x}_j^2 - \frac{k}{2} \sum_{j=1}^{N} (x_{j} - x_{j-1})^2 - m \sum_{j=0}^{N} x_j \]

we get the equations of motion:

\[ \begin{cases} \ddot{x}_{j} = \left( x_{j+1} + x_{j-1} - 2x_{j} \right) \Omega^2 - 1 & 0 < j < N \\ \ddot{x}_{0} = \left( x_1 - x_0 \right) \Omega^2 - 1 \\ \ddot{x}_{N} = \left( x_{N-1} - x_N \right) \Omega^2 - 1 \end{cases} \]

Let us eliminate gravity by writing:

\[ x_j(t) = \frac{y_j(t)}{\Omega^2} - \frac{t^2}{2} \,, \]

and then look for eigenmode solution:

\[ y_j = \sum_{k} A_{+}^{(k)} y_j^{(k)} (t) + A_{-}^{(k)} y_j^{(k)} (-t) \,,\quad y_j^{(k)}(t) = A_{jk} e^{i \omega_{k} t}\,,\]

where \( \omega_k \) are the eigenmodes' frequencies and \( A_{jk} \) the corresponding amplitude for each mass. First set aside the problem of how to find these and assume we already did. Then the initial conditions are given by these two equations:

\[ \sum_{k} A_{jk} \left( A_{+}^{(k)} + A_{-}^{(k)} \right) = \frac{(2N - j + 1)j}{2} \,,\quad \sum_{k} A_{jk} \omega_{k} \left( A_{+}^{(k)} - A_{-}^{(k)} \right) = 0 \,. \]

The eigenvector \( A_{jk} \) forms an orthonormal set. So the second equation simply gives \( A_{+}^{{k}} = A_{-}^{{k}} \). And the first equation reduces to:

\[ 2 A_{+}^{(k)} = \sum_{j} (A^{\intercal})_{kj} \frac{(2N - j + 1)j}{2} \]

Substituting this back to \( y_j \), we get the solution

\[ y_j = \sum_{k} A_{+}^{(k)} A_{jk} 2 \cos \omega_k t = \sum_{k,l} A_{jk} (A^{\intercal})_{kl} \frac{(2N - l +1)l}{2} \cos (\omega_k t) \]

We'll show below that:

\[ \begin{aligned} \omega_k^2 &= 2 \Omega^2 \left( 1 - \cos \frac{k \pi}{N+1} \right) = 4 \Omega^2 \sin^2 \frac{k \pi}{2(N+1)} \\ A_{j0} &= \frac{1}{\sqrt{N+1}} \,,\quad A_{jk} = \sqrt{\frac{2}{N+1}} \cos \left[ \left( j+\frac12 \right) \frac{k \pi}{N+1} \right] \;\; (k >0) \,, \end{aligned}\]

so that:

\[ y_{j}(t) = \sum_{k,\,l=0}^{N} \left( 2- \frac{l+1}{N+1} \right) \cos \left[ \left( j+\frac12 \right) \frac{k \pi}{N+1} \right] \cos \left[ \left( l+\frac12 \right) \frac{k \pi}{N+1} \right] \cos \left( 2 \Omega t \sin \frac{k\pi}{2(N+1)} \right) \,. \]

Here is a plot of \( x_j(t) \) for 10 particles:

The dashed trace is the motion of the center of mass. Ones sees that indeed the bottom particles hardly move until the top particle catch up with it. Of course this model allows the particle to move pass each other, which is not the case for a real slinky. We'll just ignore that.

Discussion

The discussion about compression wave having to travel all the way to the bottom is essentially correct. I got the right answer at first try but for the wrong reason. I thought that when you release the top of the slinky, the slinky wants to collapse to the center of mass. For the portion below the center of mass, there is a restoring force upward, but gravity is acting downward so the two cancels, whilst for the portion above the center of mass the restoring force and gravity are in the same direction. But it is apparent from the above figure that portion above the center of mass actually move away from the center of mass momentarily.

Perhaps my reasoning was not completely wrong after all. Consider a small time \( \epsilon \) right after release, this force of the spring is indeed equal to gravity because after all the slinky was stretched due to gravity. So the bottom part wants to collapse to the center-of-mass and wants to free fall at the same time. The two effects nicely cancel each other. To see this more concretely, consider the series expansion of \( x(t) \) in \( \epsilon \):

\[ x_j(\epsilon) - x_j(0) = y_j(\epsilon) - y_j(0) - \frac{g \epsilon^2}{2} \in O(\epsilon^{2(j+1)}) \]

So the bottom part of the slinky not only experience no force initially, but also no jerk and in fact all moments up to order \( 2j \). So all the lower order derivative of displacement are all magically zero! Well, mathematically, not magically: the sum over sinusoidal function with large phases are exactly constructively interfering. It reminds me of Feynman's interpretation of non-stationary trajectory carrying rapidly varying phases.

Appendix: Eigenmodes

The equation of motion leads to the secular equation:

\[ \begin{pmatrix} \lambda_{k} - 1 & -1 \\ -1 & \lambda_{k} & -1 \\ & \ddots & \ddots & \ddots \\ && -1 & \lambda_{k} & -1 \\ &&& -1 & \lambda_{k} - 1 & \end{pmatrix} \begin{pmatrix} A_{0k} \\ A_{1k} \\ \vdots \\ A_{(N-1)k} \\ A_{Nk} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 0 \end{pmatrix} \,, \]

where \( \lambda_k = 2 - \frac{\omega_k^2}{\Omega^2} \). The secular equation has nontrivial solution when the \( N+1 \times N+1 \) matrix \( M_{N+1} \) has zero determinant. Just apply the definition of the determinant, we have \( \det M_{N+1} = (\lambda_k -1) \det D_{N} - \det D_{N-1} \), where \( D_{N} \) is the lower right \( N \times N \) submatrix. \( \det D_{N} \) satisfies the recurrence relation \( \det D_{N} = \lambda_k \det D_{N-1} - \det D_{N-2} \) with the initial condition \( \det D_{1} = \lambda_k - 1 \) and \( \det D_{2} = \lambda_k^2 - \lambda_k - 1 \). The recurrence relation of \( \det D_{N} \) suggests the solution of the form \( \det D_{N} = a e^{\pm i b N} \). By plugging it in, we obtain the constrain:

\[ \lambda_k = 2 \cos b \,,\quad \det D_N = a_{+} e^{i b N} + a_{-} e^{-i b N} \,.\]

Using this with the boundary conditions we have:

\[ \begin{aligned} a_{+} e^{ib} + a_{-} e^{-ib} &= 2 \cos b - 1 \\ a_{+} e^{2ib} + a_{-} e^{-2ib} &= (2 \cos b)^2 - 2\cos b - 1 \end{aligned} \to\quad \begin{aligned} a_{+} & = \frac{1}{1+e^{-ib}} \\ a_{-} &= a_{+}^* = \frac{1}{1 + e^{ib}} \end{aligned} \,. \]

Therefore

\[ \det D_N = 2 \Re{\rm e} \; a_{+} e^{ib N} = \frac{\cos\left[ b \left( N + \frac{1}{2} \right) \right]}{\cos \frac{b}{2}} \,, \]

and

\[ \det M_{N+1} = (\lambda_k - 1) \det D_N - \det D_{N-1} = \frac{ \left( 2 \cos b - 1 \right) {\cos\left[ b \left(N + \frac{1}{2} \right) \right]} - {\cos\left[ b \left( N - \frac{1}{2} \right) \right]}}{\cos \frac{b}{2}} \,. \]

After a bunch of trigonometric identities, this simplifies to:

\[ \det M_{N+1} = -2 \tan\frac{b}{2} \sin b(N+1) \,. \]

The resonance frequencies are:

\[ \frac{\omega_k^2}{\Omega^2} = 2 - 2 \cos \frac{k \pi}{N+1} = 4 \sin^2 \frac{k \pi}{2(N+1)} \,, \quad k = 0,1,\dots,N\]

It is a not lot more difficult to look for \( A_{jk} \). It is a lot easier to just state the result:

\[ A_{j0} = \frac{1}{\sqrt{N+1}} \,,\quad A_{jk} = \sqrt{\frac{2}{N+1}} \cos \left[ \left( j+\frac12 \right) \frac{k \pi}{N+1} \right] \;\; (k >0) \,, \]

then check it by using the recurrence relation:

\[ - \left( A_{j-1,k} + A_{j+1,k} \right) + 2\cos \frac{k\pi}{2(N+1)} a_{j,k} = 0 \,. \]

For those of you who haven't already notice: essentially the secular matrix is diagonalized upon discrete Fourier transformation.