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Hot answers tagged coupled-oscillators

82

Your wire is not quite round (almost no wire is), and consequently it has a different vibration frequency along its principal axes1. You are exciting a mixture of the two modes of oscillation by displacing the wire along an axis that is not aligned with either of the principal axes. The subsequent motion, when analyzed along the axis of initial excitation, ...

71

As humans we oscillate left and right when we walk because we have two legs. You can get a resonance when the length of the cord is such that your pace matches the period of the swing. (Like pushing a child on a swing a little higher each time they approach you.) Whilst walking we also oscillate up and down - this can also contribute to driving the resonance....

18

I think Crawford's "Waves" is an incredible book - full of insight and clearly written by someone who loves the material. I used it for my waves-course sophomore year, and I think it's too bad it's out of print now. If you want something more theoretical, though, try Howard Georgi's book. Also, I'll second A.P.French's "Vibrations and Waves". Also, David ...

13

UPDATE : After looking again at the video, I agree that Floris' explanation seems to be correct and my explanation below is wrong. Slightly different frequencies of vibration in two perpendicular planes accounts more simply for a rotation which reverses one way then the other. Kinetic energy seems to decay constantly; it does not seem to be stored in an ...

9

Structurally, it looks like this: Start by rescaling one or both of the position/momenta pairs such that the kinetic-energy term has both masses equal (while also retaining $[x_i,p_i]=i\hbar$), and then find a rotation in the rescaled $x_1,x_2$ plane that will eliminate the coupling terms. That rigid rotation will be mirrored in the momentum plane, but ...

7

The origin of the mass matrix lies in a change of coordinates from Cartesian to generalized coordinates. The kinetic energy of a system of $N$ particles in terms of Cartesian coordinates is $$K=\frac 12\sum_{a=1}^Nm_a\dot{\vec r}_a\cdot\dot{\vec r}_a.$$ If the system is scleronomic, i.e., the relation between Cartesian and generalized coordinates do not ...

6

On the top of Figure we have $\:n+1\:$ ideal springs and $\:n\:$ particles in equilibrium. The constants of the springs are $\:k_{\rho}\: (\rho=1,2,\cdots, n+1) \:$ with equilibrium lengths $\:\ell_{\rho}\:(\rho=1,2,\cdots, n+1 )\:$ and the particle masses $\:m_{\rho}\:(\rho=1,2,\cdots, n)$. Disturbing the system from this equilibrium, the equation ...

6

An engineer would call it the "mass matrix" not the "kinetic energy matrix". The KE is given by $\frac 1 2 \mathbf{v}^T \mathbf{M} \mathbf{v}$ where $\mathbf{v}$ is the vector of the velocity components $\dot x_1, \dots, \dot x_4, \dot y_1, \dots, \dot y_4$ and $\mathbf{M} = M \mathbf{I}_8$ - i.e. an $8\times8$ diagonal matrix with all the diagonal terms ...

5

First of all I guess that what you wrote is the Hamiltonian and not the Lagrangian of the system and $\dot{x}$ stays for $p_x$ and $\dot{y}$ stays for $p_y$. You can decouple the problem redefining $$(X,Y)^t = R(x,y)^t$$ for a suitable $R\in O(2)$ diagonalizing the symmetric matrix in the potential part of your Hamiltonian. This way you see the final ...

5

The result $\omega^2=\frac{c_1+c_2}{M} \pm \frac{1}{M}\sqrt{c_1^2+c_2^2+2c_1c_2 \cos ka}$ leads to two real solutions for $\omega^2$, since $-1 \lt \cos ka \lt 1$ and the square root lies between $|c_1 - c_2|$ (for $\cos ka = -1$) and $c_1 + c_2$ (for $\cos ka = +1$), so that $\omega^2$ is always positive. The frequency $\omega$ is taken as a positive ...

5

The Hamiltonian is a quadratic form both in the momenta and in the coordinates. You can decouple the problem by using the principal axis theorem on the coordinates to obtain generalized coordinates. For this, first a scaling of the coordinates to equalize the coefficients of the kinetic energy is necessary. The whole approach corresponds to "diagonalizing" ...

4

This is actually an interesting problem in classical mechanics, dating back to Huygens. We'll work with the three variables you define in the question, namely $(x,\theta_1, \theta_2)$. Also we'll set your $m = l = g = 1$ for simplicity. Kinetic Energy The position vector of the first pendulum bob is $$\mathbb{r}_1 = (x+\sin\theta_1, \cos\theta_1)$$ ...

4

We consider any potential of the form \begin{align} V( x) = K_{11}x_1^2 + 2K_{12} x_1x_2 + K_{22}x_2^2. \end{align} The key to "diagonalization" of this potential is to note that such a potential is quadratic in the positions $x_1$ and $x_2$, and can therefore be written in matrix notation as follows: \begin{align} V( x) = \underbrace{(x_1\, x_2)}_{ x^t}\...

4

The best thing to do is to look at the eigenvectors. This will tell you all you need to know. Anyway, $0$ corresponds to all the masses rotating in unison. Another mode corresponds to neighbouring masses moving closer and farther away from each other, symmetrically. Here you can choose a generalized coordinate $x$ to be the distance between one pair of ...

4

Yes. Consider electromagnetic fields in a square box of volume $V = L^3$. One can then impose periodic boundary conditions on the vector potential, and one arrives at the following expansion in terms of Fourier modes: \begin{align} \mathbf A(t,\mathbf x) = \sum_\mathbf k\sum_r \left(\frac{\hbar c^2}{2V\omega_\mathbf k}\right)^{1/2} \mathbf \epsilon_r(\...

4

1 What does potential mean?(I know that potential function defines an invariance most of the times, but how it is defnied for oscillators case? Here it should be defined as the potential energy of the system. This page should clarify a little bit wikipedia entry on potential energy. In general potentials are quite a complex topic, but here it should suffice ...

4

If the eigienvalues form a continuous spectrum, like the eigenvalues of $x$, then states must be normalized to a dirac delta, $$\left\langle x \right| x' \rangle = \delta(x-x')$$ The trace of an operator is the sum of the diagonal elements, or if the basis is continuous, it becomes an integral \begin{eqnarray*} \mathrm{Tr}\left(\left|\phi\right\rangle\...

4

You are making to the following mistakes: Your initial displacement $d_0$ is 5 % of the variation of the respective variable, while it should be orders of magnitude lower. The Lyapunov exponent is defined for the limit of infinitesimal displacements, i.e., $d_0→0$. Your observation time is far too short. You are looking at four oscillations of your dynamics,...

4

The equation that dictates the behavior of mechanical systems such as oscillators is Newton's Second Law. It is a second order differential equation, and in the case of oscillators it is linear and has the general form $$\ddot{\vec{x}} = A\vec{x} .$$ This means that for each degree of freedom you have, another coordinate will be assigned to it and the ...

4

I just add to above points which are mostly correct my two pennies. 1- the wire is loaded with a history of residue strains from the manufacturing and handling so the stiffness and elasticity of it is not homogenous lengthwise or even along cross-section. if you'd write the whiplash DE's in a finite element software it is not a linear differential equation ...

4

My first instinct as a physicist is always to ask whether symmetry impacts you at all. From my understanding, you had three ions spaced in a quadratic potential (and, I assume, interacting with each other). It seems clear to me that you must have started in a configuration like so: | o --- o --- o | That is, the ions would have been arranged ...

4

It's a special case of $$H = \sum_i \frac{P_i^2}{2m} + \sum_{ij} Q_{ij} X_i X_j$$ where $Q_{ij}$ is a real symmetric matrix. Since $Q_{ij}$ is orthogonally diagonisable, the system is equivalent to a set of uncoupled harmonic oscillators. In principle, everything you could want to compute can then be expressed in terms of the eigenvalues and eigenvectors ...

4

Yes, there is. The keywords you're looking for are collective behavior and, in particular, synchronization in dynamical systems. And yes again: there must be some sort of coupling and, in a discrete model, the coupling between the individual oscillators will typically take the form of a synchronization network. A recent (2015) review is Synchronization of ...

4

I suppose the Lagrangian you found is actually $$L = \frac{1}{2} \left( \dot{q}_1^2 + \dot{q}_2^2 \right) - \frac{q_1^3}{3} + q_1 q_2^2$$ Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find $$\ddot{q}_2 - 2 q_1 q_2 = 0$$ There's no way that I can see to change the sign of the second term without affecting the ...

4

Hint: $$x^2+y^2+xy = \frac34 (x+y)^2 + \frac14 (x-y)^2.$$ This means that if you transform to the new variables \begin{align} \xi & = \frac1{\sqrt{2}}(x+y) \\ \eta & = \frac1{\sqrt{2}}(x-y), \end{align} and with a similar transformation from $p_x,p_y$ to $p_\xi,p_\eta$ to make sure that $[\xi,p_\xi] = i = [\eta,p_\eta]$ and [\xi,p_\eta] = 0 = [\... 3 From the free body diagram you must have \begin{align} m_1 \ddot{x}_1 &= F_1 - F_2 \\ m_2 \ddot{x}_2 &= F_2 - F_3 \\ m_2 \ddot{x}_3 &= F_3 \end {align} with the spring forces defined as \begin{align} F_1 & = -k_1 x_1 \\ F_2 & = -k_2 (x_2-x_1) \\ F_3 & = -k_3 (x_3-x_1) \end{align} The above is combined as\$ \begin{...

3

What it is trying to say is that white noise is uncorrelated (truly random, implying a zero correlation time) while real noise is not actually random and has a pattern to it that repeats in finite time. So the correlation time it is referring to is the amount of time for the signal to "repeat" in a statistical sense. I don't see a relaxation time in the ...

3

I highly recommend Waves and Oscillations. A Prelude to Quantum Mechanics. It takes a Physics perspective, which may or may not be what you want. I like that it focuses more on physical insight rather than mathematical rigor. The later chapters are a great introduction to Quantum Mechanics.

3

I think you should try these books: 1) David Tong: Lectures on Classical Dynamics (http://www.damtp.cam.ac.uk/user/tong/dynamics.htm) 2) V. I. Arnold, Mathematical Methods of Classical Mechanics 3) L. Landau an E. Lifshitz, Mechanics

3

The zero frequency mode corresponds to a merry-go-round of masses rotating in the same direction. No springs are being stretched/compressed so there is nothing (apart from friction of course) to stop and reverse the motion of the masses.

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