# Tag Info

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Chaotic is not the same as random. A chaotic system is entirely deterministic, while a random system is entirely non-deterministic. Chaotic means that infinitesimally close initial conditions lead to arbitrarily large divergences as the system evolves. But it's impossible, practically speaking, to reproduce the same initial conditions twice. Given ...

17

The effect of gravity is miniscule, and here's why: The speed of sound in a string is basically $$v = \sqrt{\frac{T}{\lambda}},$$ where $T$ is the tension and $\lambda = M/L$ is the mass per unit length. The frequency of a plucked string will then be this sound divided by the length of the oscillator: $f = v/L$. Combining and rearranging tells us the ...

10

These kinds of proportionality questions are often best answered with dimensional analysis. You want to know a form a quantity with the units of time in terms of what you have. You have a quantity $k$ with units $\frac{\text{Energy}}{\text{Distance}^3} = \frac{\text{Mass}}{\text{Distance} \times \text{Time}^2}$. You also have the mass $m$ (units of Mass) ...

7

Sometimes, a good figure is worth more than a thousand equations :) I numerically integrated the following equation of motion for a physical pendulum: $$I\ddot{\theta} - mgL\sin(\theta) + \frac12\mathrm{sgn}({\dot{\theta}})L\rho_{\mathrm{air}}C_DS(L\dot{\theta})^2 - \gamma\theta = 0$$ with $\mathrm{sgn()}$ the signum function. The second term in this ...

6

It would depend on damping effects being taken into account or not. Invoking Newton's 2nd Law of motion, a differential equation for the motion of a damped harmonic oscillator can be written (including an external, sinusoidal driving force term): $m\frac{d^2x}{dt^2}+2m\xi\omega_0\frac{dx}{dt}+m\omega_0^2x=F_0\sin\left(\omega t\right)$ Where $m$ is the ...

4

The principle is the same in both pictures. I'm not sure how to answer "is this analogy significant?", since I don't think it's an analogy at all. It's the same phenomena, as explained below. Any time you have a mechanism for dissipation (a coupling between the "system" and a "bath") that coupling mechanism will give rise to back-action of the bath the ...

4

The vast majority of research cyclotrons don't use the classical Lawrence design anymore. Lawrence's cyclotron was in many ways the simplest circular accelerator you can build, and later designs are much more complex. That said, we can still consider a classical cyclotron with four accelerating gaps. To review a bit, the cyclotron uses a perpendicular ...

4

No, it is not. Your system will go through the same point twice in every oscillation, once moving in each direction, and the friction force will be reversed in each pass, so your approach doesn't work. What you need to consider is the velocity, not the displacement, so $$ma=-kx - \mathrm{sign}(v) F_{\mathrm{fric}}.$$ This is not all that helpful in ...

4

This is really the same as Mark's answer but phrased in a different way. If you write your equation in Leibniz's notation it is: $$\frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\sin(\theta)$$ This makes it clearer that the solution is going to be the function $\theta(t)$ i.e. the angle as a function of time.

4

The dimensional analysis in zkf's answer completely solves the exercise. Nevertheless, it is possible to give a closed formula for the period $$T~=~ 4 ~\sqrt{\frac{m}{2k}} \int_0^a\! \frac{dx}{\sqrt{a^3-x^3}} ~\stackrel{x=au}{=}~ 4 ~\sqrt{\frac{m}{2ka}} \int_0^1\! \frac{du}{\sqrt{1-u^3}}.$$ Can you see why? Unsurprisingly, this just confirms zkf's ...

4

zkf gives you enough to answer this question but I would like to make a few extra points: The absolute value operation in the potential makes this a nonlinear problem, which are generally pretty difficult to deal with. I got impatient waiting for Mathematica to come up with a closed form solution for $x(t)$, so there probably isn't one. This is the typical ...

4

Yes, principally because the speed of sound depends on the temperature. An approximate equation for the speed of sound in dry air is: $$v = 331 + 0.6T$$ The wavelength is fixed by the pipe length so if the speed of sound changes the frequency also changes according to: $$f = \frac{v}{\lambda}$$ In principle there will be some thermal expansion of the ...

4

So I am not an expert in limit cycles by any means but I am intrigued by this problem so here is what I came up with. Let's treat the nonlinear term perturbatively. This will not be enough to prove the existence of the limit cycle for large values of $\mu$, but given that apparently there is a proof that this works perturbatively, it will be enough for us. ...

4

I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian $L=L(q,\dot{q})$ of $n$ generalized coordinates $$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$ One of the first questions one would like to ask is, if a specific ...

3

If you solve the equation for $\theta$, whether numerically or otherwise, you will have $\theta$, not $\dot\theta$ or $\ddot{\theta}$. You can get those by differentiating, of course, or you may find them as part of your numerical algorithm along the way. The fact that the equation is second order simply means you will need two initial conditions to find a ...

3

The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force ...

3

The phase constant is needed only if you have a specific initial condition, e.g. if I told you where $x$ was at time $t = 0$, you could solve for $\varphi$. Otherwise you can just choose whatever you want for it: Note that it is the same in all functions. Choosing some value for $\varphi$ is analogous to you manually setting the time origin to something ...

3

How about instead of finding the roots and then making the plots, you skip right to the plots using a Monte Carlo method? Choose a random k, then choose a random ω and calculate the left-hand side and the right-hand side of the R-L equations. If the RHS is close enough to the LHS (you pick how close), put a point on the plot (blue or black, depending ...

3

The reason for this result is the sign in you damping term. For a damped harmonic oscillator you need to have a resistive force on the mass point at $x$. That means if $x=0$ is the equilibrium position the damping term will be proportional to the velocity with an negative constant $F_{\text{damp}} = -ax',a>0$. I.e. the total force on the mass point ...

3

The inductor and capacitor form a resonant circuit, which will pass only a specific frequency - the one you are tuning the radio to recieve. You normally tune it by making either the inductor or capacitor adjustable. edit: As described in How does radio receives signal from particular station? it's very much like a pendulum. Current flows freely in the ...

3

$x_{max}$ is the amplitude of the oscillations, and yes, ${\omega}t - \varphi$ is the phase. We know that the period $T$, is the reciprocal of the frequency $f$, or $$T = 1/f$$ We also know that $\omega$, the angular frequency, is equal to $2\pi$ times the frequency, or $$\omega = 2{\pi}f$$ From here, we can use the initial conditions to find the ...

3

The period of the pendulum is roughly $T\approx2\pi\sqrt{\frac{L}{F}}$ Where $L$ is its length and $F$ is the downward-pulling force (usually gravity). Let's call our reference period $T_{rest}$. Now let's examine your cases. 1. When the train is in circular motion in a curve of radius R with constant speed. If the train is in circular motion the ...

3

The differential $dp(x)$ is the probability of finding the body in an interval of length $dx$ centered at $x$. The quantity $p$ you are looking for is the cumulative distribution function, $$P(x)=\int_{-\infty}^x \frac{dp}{dx}(x) dx,$$ which is the probability that the particle will be to the left of the point $x$. Since the particle cannot be to the left of ...

3

Does it just mean that to a small approximation a body will not oscillate? It means that you must always remember the context in which a formula is valid and not "blindly" apply it. Where does the formula come from? Consider the homogeneous differential equation for the harmonic oscillator: $$\ddot x + \dfrac{k}{m}x = 0$$ with solutions $$x(t) = ... 3 When you say "Suppose I boost myself to the rest frame" you have in effect asserted that you are looking at a particular mass state rather than at the superposition of mass-states that arises from allowing a flavor state to propagate freely. This is related to the question of how neutrinos can oscillate though the lepton flavors have differing masses (also ... 3 Perhaps a better question to ask is: why is a single pendulum non-chaotic? Almost all real systems are chaotic at least to some extent; the fact that we can write out the solution for a single pendulum for all points in time is really quite peculiar, and only true because it is a highly simplified system. The reason these non-chaotic systems are so prevalent ... 2 First, the fact that your graph goes to negative infinity would indeed mean that the spring will break, or at least that it's going to stop behaving like a spring. Basically, consider that you have obtained this solution using a model of a spring in which the force it exerts is proportional to its displacement from equilibrium. Once you reach the point at ... 2 Nibot, I strongly suggest you read Noise and Fluctuations by D.K.C. MacDonald. It has lots of great discussions related to thermal noise. That's where most of this answer comes from. You are probably used to the Fluctuation Dissipation Theorem (FDT) written in a form similar to the way Nyquist derived Johnson noise on a resistor:$$ <\delta V_f^2> = ...

2

The easiest way to determine maximum and minumums of a function is to set the derivative equal to zero. Thus, in this case, setting the equation for acceleration equal to zero and solving for the variables of interest will give you what you want. Thus, in this case we have the equation for position: $$x = a \sin(\omega t + \phi)$$ One way to see the ...

2

Do you know what the solutions to the equation of motion look like in the case where there's only one sinusoidal term on the right-hand side, i.e., $$\ddot z-\omega^2z={f_0\over 2}e^{-i(\omega+\delta)t},$$ and the corresponding equation with $\omega-\delta$? If so, then you're most of the way there. When you have an inhomogeneous linear differential ...

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