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I would like to understand the behaviour of a simple mass-and-spring system - a classical harmonic oscillator - in the $xy$ plane that is in rotation about $\hat z$ with frequency $\vec \Omega=\Omega\hat z$. In connection with the figure below:

enter image description here

I just imagine that all I do is set the frame in rotational motion in the plane, without giving the frame any initial translation but initially displacing the mass along $x$ only.

I'm interested in the regime where the $y$-springs are very weak, i.e. $k_y\approx 0$, but I will also compare with the regime where the springs in the $\hat y$ direction have the same spring constants as those in the $\hat x$ direction: $k_y=k_x$. I always use as initial conditions that the mass is launched from rest from some position $X_0> 0$ and $Y_0=0$. Using dimensionless time $\tau =\sqrt{\frac{k_x}{m}}t =\omega t$ and using ${m\omega^2 X_0^2}$ as the energy scale it is easy to construct the dimensionless Lagrangian \begin{align} L= \frac{1}{2}\left(\dot{x}^2+\dot{y}^2\right) +\frac{1}{2}r^2(x^2+y^2)+ r \left(x\dot{y}-y\dot{x}\right)-2\times \frac{1}{2}\left(x^2+\eta y^2\right) \end{align} where \begin{align} x=\frac{X}{X_0}, \quad y=\frac{Y}{X_0}\, ,\qquad r=\frac{\Omega}{\omega}\, ,\qquad \eta=\frac{k_y}{k_x} \, , \end{align} are all dimensionless, and $\dot{x}=dx/d\tau$ etc.
The equations of motions follow immediately: \begin{align} \ddot{y}&=-2r\dot{x} +(r^2-2\eta)y \\ \ddot{x}&= 2r\dot{y}-(2-r^2)x\, . \tag{3} \end{align}

I will assume $r<1$, i.e. the angular rotation $\Omega$ about $\hat z$ is always slower than the angular frequency $\omega$ of the $x$-oscillator.

It seems the qualitative features of the motion depend on whether $(r^2-2\eta) > 1$ or $<1$ in Eq.(3). This is possibly surprising as simple "power counting" arguments indicate that the $2r\dot{x}$, which has only one power of the small parameter $r$, is the one that should dominate. I want to better understand why the $2r\dot{x}$ term has apparently little to no effect on the motion, and understand any reasonably rigorous path to an analytical solution.

For the case where $\eta=1$, $r^2-2\eta <1$ and the problem in the $r\ll 1$ regime is basically identical to the Foucault pendulum. Dropping terms in $r^2$ one can solve analytically by introducing $s=x+iy$. The solution can be found in many textbooks, such as Landau and Lifshitz. For completeness, some elements of (numerical) solutions are found at the end of the post.

In the case where $r^2-2\eta >1$, which would be the case when the $y$-springs are removed or at least are much weaker than the $x$-springs, the solution is apparently that of an inverted oscillator. For $r=0.1$ and $\eta=1/150$, the solution shows that $y$ constantly increases with some small oscillations; the average $x$ motion also drifts from $\langle x\rangle=0$ for small $\tau$ to $\langle x\rangle>0$ for later times.

On the left below, the red curve is the log of the $y(\tau)$-position; the black curve is $x(\tau)$.

enter image description here enter image description here

For long times, it clear that the solution in $y$ is exponential with $\tau$, which goes against the naive suggestion that the equation of motion for $y$ in Eq.(3) should be, for small $r$, dominated by the term in $-2r \dot{x}$. In other words, a naive perturbative approach where \begin{align} y(\tau)&= y^{(0)}(\tau)+ ry^{(1)}(\tau)+r^2 y^{(2)}(\tau)\\ x(\tau)&= x^{(0)}(\tau)+ rx^{(1)}(\tau)+r^2 x^{(2)}(\tau) \end{align} doesn't seem promising since, to order $r$ and for $\eta\sim {\cal O}(r^2)$, one would have \begin{align} y^{(0)}(\tau)=0\, ,\qquad \ddot{y}^{(1)}(\tau)=2\sqrt{2} r \sin \sqrt{2}\tau -2\eta y^{(1)}(\tau) \end{align} with $x^{(0)}(\tau)=\cos \sqrt{2}\tau$ satisfying the initial conditions. This looks like a driven oscillator but, if I choose $\eta\sim r^2$ and ignore this term as much smaller, I get $y^{(1)}(\tau)\sim \sqrt{2} r\sin(\sqrt{2}\tau)$, which satisfies the boundary conditions on $y$ to order $r$ but is numerically false: $y(\tau)$ for small $\tau$ is flat until $\tau\approx \pi/2$ and then rapidly rises without crossing the $y$-axis.

Since $L$ does not depend explicitly on time, $H$ is conserved and, as a sanity check I have verified that the dimensionless $H$ remains constant (to within numerical errors). Indeed, it varies by at most $8$ parts in 500000, so the numerical accuracy of the solution is not an immediate concern. Given that the potential energy of the spring $\sim x^2$, the stretching of the spring can be well beyond the initial condition $x(0)=1$ but the Hamiltonian in the rotating frame is $H=H_0-\Omega \ell_z$ (here dimensionful form, with $\ell_z$ the angular momentum) so in principle the spring system could exceed its energy in the inertial frame and oscillate in $x$ with considerable amplitude.

Q: Is there a way to understand the large $\tau$ behaviour in the case $r^2-\eta >1$ using simple physics arguments, and possibly analytically? Afterall, the case $\eta=0$ corresponds to no spring along $y$ and it doesn't seem very intuitive that the oscillator should not stay near the origin in $y$, especially that it stays near the origin when the $y$-springs are present.

Mathematically, can one legitimately "average" the $\dot{x}(t)$ in (3) (this motion is fast compared to the slow rotation), either subbing directly $\langle \dot x(\tau)\rangle\approx 0$ to get an inverted oscillator or perhaps more formally importing the argument of the Kapitza pendulum to produce an effective potential that would include this average? Note that numerically, $\langle \dot{x}\rangle \approx 0$ even for a cycle close to $\tau=100$.

I have no explanation for the drift in $x$ away from the origin.


Additional material:

The two figures below are: on the left the $x$ (in black) and $y$ (in red) positions as a function of time; on the right is a parametric plot $(x(t),y(t))$. This is for $\eta=1$ and $r=1/10$ so that $r^2-2\eta=-1.99$. One can see that the plane vertical plane of oscillation of the pendulum slowly rotates about $\hat z$, as in the Foucault pendulum. The exact same numerical integrator is used for both cases of the sign of $r^2-2\eta$.

enter image description here enter image description here

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  • $\begingroup$ Equation (Eq. (3)) of motion seems simple coupled second order linear ordinary differential equations? Hence it is straight forward to solve it through Laplace transform. Is it not the analytical solution which is asked for? Is $\Omega$ dependent on time? $\endgroup$ – Sunyam Feb 11 at 14:12
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    $\begingroup$ @Sunyam Thanks for your comment. $\Omega$ is independent of time but the problem is with the physics: even though the physically dominant term is due to the spring and the rotation should be a perturbation, it is clearly not so I'm missing some physics to understand the behaviour of the solution. I haven't looked at this in a while so maybe I should just come back to this sooner rather than later. $\endgroup$ – ZeroTheHero Feb 11 at 18:20

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