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## Hot answers tagged harmonic-oscillator

5

Ach, yes & no, but... this is the most ponderous summary of the classic Jordan map construction of SU(2) matrix generators there is. Most of the difficulty and confusion here lies in the change of bases, which runaway abstraction of notation fosters. The notional spaces of the two oscillators are not the spacetime indexed by the generators of rotations, ...

4

The period is independent of the amplitude keeping $k$ and $m$ fixed (and varying initial conditions). Your calculation tries to keep the initial conditions fixed and vary $k$ and $m.$

2

Since the derivation from your question is a bit rough, I take the liberty to elaborate on it a little more. The first equation from your question can be written in the form $$\frac{d}{dt}\left( ml^2 \dot{\phi}\right) + mgl\sin(\phi)=0$$ Trying to find the corresponding Lagrangian, we can further write this as $$\frac{d}{dt}\frac{\partial}{\partial \dot \phi}... 2 No, that would not be sensible. Consistency requires that as t\to 0, K(x_b,t_b,x_a,t_a) \to \delta(x_b-x_a). And indeed this is true. In the limit of small t, the propagator becomes$$ K(x_b,x_a;t) \simeq \sqrt{\frac{m}{2\pi it}}\exp\left\{-\frac{m}{2it}(x_b-x_a)^2\right\} $$which by itself is perfectly divergent as you say. It is however a well-known ... 2 What do the units of 1/acceleration mean? We can understand it in terms of the units of acceleration. Write$$ 15~\frac{\textrm{m}}{\textrm{s}^2} $$as$$ \frac{15~\textrm{m}/\textrm{s}}{\textrm{s}}, or, in words, "15 (meter per second), per second". Which is to say, the velocity is changing by 15 m/s every second. Thinking about this in a ... 1 In a sinusoidaly forced RLC circuit, the maximum charge and energy on the capacitor occurs at a frequency just below resonance. The current changes very little, but there is more time on each cycle to charge the capacitor. The maximum voltage and energy on the inductor occurs just above resonance. The rate of change of the current is a little higher. The ... 1 The error was that I used \epsilon = t_b - t_a instead of \epsilon = (t_b-t_a)/2. The kernel works now. 1 Indeed all the particles have the same amplitude of vibration in both transverse and longitudinal waves. The exception being standing waves, which are a combination of waves. Standing waves produce nodes and anti-nodes, nodes being point of zero amplitude, and anti-nodes being points of maximum amplitude. The equation of a standing wave is as follows (it is ... 1 You were correctly looking at the general form of the response of a harmonic oscillator, where you get the \frac{1}{\omega_n^2-\omega^2} proportionality term between the displacement and the force (I omit some constants here). From here, if \omega_n\ll\omega, you get a minus sign. So the force and displacement are just like in the 1st figure, at a 180 ... 1 I'm not sure why you're making this so complicated. Write \begin{align} x(t)=A\cos(\omega t)+B\sin(\omega t) \, ,\tag{1} \end{align} where \omega^2=k/m, A=x(0) and B= \dot{x}(0)/\omega. Since x(0) and \dot{x}(0) are arbitrary, so are A and B. Now rewrite (1) as \begin{align} x(t)=C\cos(\omega t-\varphi)=C\cos(\varphi)\cos(\omega t)+C\sin(\... 1 It is just related to two different initial conditions for the system. Assume the system is a mass attached to a spring. Ifx=A\cos(\omega t)$$then this represents holding the mass with the spring extended in the +x direction at t=0. When you let go, the velocity is in the negative x direction. This agrees with the calculated velocity you get by taking ... 1 The combined system eigenstate is not a linear combination. When you combine two quantum systems you take a tensor product of the Hilbert spaces so the combined eigenstates are$$ |n_1,n_2\rangle \equiv |n_1\rangle \otimes|n_2\rangle $$where the first form is the way that a physicist would write it, and the second the way a methematician would write it. ... 1 The statement is poorly worded, but it must mean always having the same speed, rather than just momentarily. Two particles that are separated by a distance of k\lambda will have a phase difference of 2 \pi k. But if the particles always have the same speed then their phase difference must be a multiple of \pi. (Why ? Suppose the phase difference is, ... 1 We can see from the exponential term e^{-\frac{b}{2m}t} that the amplitude in this model of the pendulum’s motion never actually reaches zero. It is possible to ask how long it takes the amplitude to reach some small value - say one hundredth or one thousandth of A_0 - but it is not meaningful to ask how long it takes the pendulum to stop completely. Of ... 1 The order depends on where you drop the terms. You have two levels of description, the Lagrangian or Hamiltonian of your system and the equation of motion derived from it. The harmonic approximation at the level of the Lagrangian (or Hamiltonian) is exactly that: dropping all terms beyond quadratic order. So you have a second order polynomial as your ... 1 The equations of motion are:$${\ddot l}-{\dot\varphi }^{2}l-g\cos \left( \varphi \right)+\frac km\,l=0\\ \ddot\varphi +2\,{\frac {{\dot l}\,\dot\varphi }{l}}+{\frac {g\sin \left( \varphi \right) }{l}}=0 $$Linearization$$\cos(\varphi)=1-\frac 12\,\varphi^2 +\frac{1}{24}\varphi^4+\ldots\approx 1\\ \sin(\varphi)=\varphi-\frac 16 \varphi^3+\frac{1}{120} \...

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