# Tag Info

## Hot answers tagged oscillators

39

The balloon has a very small mass and friction is large (large surface area), so the oscillation is very damped.

26

Part of it is that since Newtonian mechanics is described in terms of calculus. When we consider vibrational motions, we're talking about some particle that tends to not be displaced from some equilibrium position. That is, the force on the particle, at displacement $x$, $F(x)$, is equal to some function of displacement $x$, $g(x)$. There are two ways ...

22

Tied balloons do behave like a pendulum, you only need really massive ones: You can see it live in a video. Hot air balloons have such a big amount of..erm...hot air that during the start you can expect oscillations because while the surface area is big, the mass inside is so big that the dampening is low enough. They are not exactly like a pendulum ...

20

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 ...

13

One of the big reasons not discussed above is Fourier theory -- any function $f(x)$ can be expressed in the form $f(x) = \int dk\, A(k)e^{ikx}$, which basically means that any function can be decomposed into an infinite sum of sines and cosines. Since this is the case and dealing with sine and cosine is mathematically simpler than the general case of ...

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) ...

9

Because cycles and oscillations and things with periodicity, are all intimately related to the circle. And $sin$ and $cos$ are defined based on the circle.

8

A literal response to your title question would simply be "because in the physical world, oscillations behave in ways consistent with $\sin$ and $\cos$." Of course, one then wonders why these functions are so ubiquitous. Depending on your level of physics background, you may be familiar with the harmonic oscillator - that is, a system for which there exists ...

7

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

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 ...

6

Actually, it does behave exactly like a pendulum. The equations of motion are exactly the same. The issue, as Jasper pointed out, is damping. When you think of a "normal" pendulum, you are considering a lightly damped oscillator. The balloon, as I will prove below, is heavily damped. For a damped harmonic oscillator (which a pendulum approaches for small ...

5

You almost answered it on your own! Essentially it's the ratio of the viscous force to the gravitational force. As $\beta \rightarrow 0$, the gravitational force dominates and the damping due to air friction is very small. Likewise, as $\beta \rightarrow \infty$, the air friction dominates the solution. This isn't really all that illustrative physically, ...

4

What you wrote is a solution, it is just not the only form of the solution. Here are a couple ways to proceed: What happens when $4km < b^2$ ? Let's march ahead done this road. The value in the square root becomes negative, and so there is no real solution for $\omega$. However you can find imaginary solutions for $\omega$. $$\omega = \frac{\pm ... 4 No, you're not missing the obvious: this is a good question. The simple harmonic oscillator shoved down your throat in freshman physics and engineering courses is a linear system: any solution scaled by a scale factor is also a solution. So the period cannot depend on amplitude. The basic equation defining this beast is$$\ddot{x} = - \omega^2 x\tag{1}$$... 4 1 What does potential mean?(I know that potential function defines an invariance most of the times, but how it is defnied for oscillators case? Here it should be defined as the potential energy of the system. This page should clarify a little bit wikipedia entry on potential energy. In general potentials are quite a complex topic, but here it should ... 4 No, it is not. Your system will go through the same point twice in every oscillation, once moving in each direction, and the friction force will be reversed in each pass, so your approach doesn't work. What you need to consider is the velocity, not the displacement, so$$ma=-kx - \mathrm{sign}(v) F_{\mathrm{fric}}.$$This is not all that helpful in ... 4 The principle is the same in both pictures. I'm not sure how to answer "is this analogy significant?", since I don't think it's an analogy at all. It's the same phenomena, as explained below. Any time you have a mechanism for dissipation (a coupling between the "system" and a "bath") that coupling mechanism will give rise to back-action of the bath the ... 4 Yes, principally because the speed of sound depends on the temperature. An approximate equation for the speed of sound in dry air is:$$ 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} $$In principle there will be some thermal expansion of the ... 4 The dimensional analysis in zkf's answer completely solves the exercise. Nevertheless, it is possible to give a closed formula for the period$$ 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}}. $$Can you see why? Unsurprisingly, this just confirms zkf's ... 4 zkf gives you enough to answer this question but I would like to make a few extra points: The absolute value operation in the potential makes this a nonlinear problem, which are generally pretty difficult to deal with. I got impatient waiting for Mathematica to come up with a closed form solution for x(t), so there probably isn't one. This is the typical ... 4 The phase constant is needed only if you have a specific initial condition, e.g. if I told you where x was at time t = 0, you could solve for \varphi. Otherwise you can just choose whatever you want for it: Note that it is the same in all functions. Choosing some value for \varphi is analogous to you manually setting the time origin to something ... 4 The vast majority of research cyclotrons don't use the classical Lawrence design anymore. Lawrence's cyclotron was in many ways the simplest circular accelerator you can build, and later designs are much more complex. That said, we can still consider a classical cyclotron with four accelerating gaps. To review a bit, the cyclotron uses a perpendicular ... 4 This is really the same as Mark's answer but phrased in a different way. If you write your equation in Leibniz's notation it is:$$ \frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\sin(\theta) $$This makes it clearer that the solution is going to be the function \theta(t) i.e. the angle as a function of time. 4 I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian L=L(q,\dot{q}) of n generalized coordinates$$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$One of the first questions one would like to ask is, if a specific ... 3 Does it just mean that to a small approximation a body will not oscillate? It means that you must always remember the context in which a formula is valid and not "blindly" apply it. Where does the formula come from? Consider the homogeneous differential equation for the harmonic oscillator:$$\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

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

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. ...

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 ...

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