# Tag Info

39

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

30

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 ...

18

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 ...

16

Usually, the (sinusoidal) driven harmonic oscillator is damped, and the first two parts of your solution (which depend on the initial conditions, while the third term does not) are transient, i.e. not relevant after a short time. That the solution $$x(t) = \frac{F_0\sin(\omega t)}{m(\omega_0^2 - \omega^2)}$$ cannot be the "full" solution to the equation of ...

14

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 ...

13

There seem to be a lot of human body mechanical models, such as this one: As for applications, I have heard that sub-audio frequency vibrations have been considered as nonlethal weapons for riot control.

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 ...

8

In a very mathematical sense, more often than not a mode refers to an eigenvector of a linear equation. Consider the coupled springs problem $$\frac{d}{dt^2} \left[ \begin{array}{cc} x_1 \\ x_2 \end{array} \right] =\left[ \begin{array}{cc} - 2 \omega_0^2 & \omega_0^2 \\ \omega_0^2 & - \omega_0^2 \end{array} \right] \left[ \begin{array}{cc} x_1 \\ x_2 ... 7 It would depend on damping effects being taken into account or not. Invoking Newton's 2nd Law of motion, a differential equation for the motion of a damped harmonic oscillator can be written (including an external, sinusoidal driving force term): m\frac{d^2x}{dt^2}+2m\xi\omega_0\frac{dx}{dt}+m\omega_0^2x=F_0\sin\left(\omega t\right) Where m is the ... 7 Sometimes, a good figure is worth more than a thousand equations :) I numerically integrated the following equation of motion for a physical pendulum:$$ 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 $$with \mathrm{sgn()} the signum function. The ... 6 I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian L=L(q,\dot{q}) of n generalized coordinates$$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$One of the first questions one would like to ask is, if a specific ... 6 Actually, it does behave exactly like a pendulum. The equations of motion are exactly the same. The issue, as Jasper pointed out, is damping. When you think of a "normal" pendulum, you are considering a lightly damped oscillator. The balloon, as I will prove below, is heavily damped. For a damped harmonic oscillator (which a pendulum approaches for small ... 6 In refraction and reflection the incoming electromagnetic wave causes the electron density of the refracting material to oscillate. This happens because at any point in space the wave produces an oscillating electric field (and magnetic field, though that isn't relevant here) so any material that has a non-zero polarisability will respond by developing an ... 6 Maybe the pendulum would be an interesting example. The equation of motion of an ideal pendulum of length \ell in a uniform gravity field is$${d^2\theta\over dt^2}+{g\over \ell} \sin\theta=0$$It is well known that in the small angle approximation this reduces to a harmonic oscillator, but the precise solution involves elliptic integrals, and can be ... 5 This is an answer by an experimentalist who had been fitting data with mathematical models since 1968. When fitting data one goes to the simplest mathematical models. When the data display variations in time and space the Fourier expansion is extremely useful because it gives the frequencies and amplitudes that will fit a periodic data set. One gets as ... 5 You almost answered it on your own! Essentially it's the ratio of the viscous force to the gravitational force. As \beta \rightarrow 0, the gravitational force dominates and the damping due to air friction is very small. Likewise, as \beta \rightarrow \infty, the air friction dominates the solution. This isn't really all that illustrative physically, ... 5 The principle is the same in both pictures. I'm not sure how to answer "is this analogy significant?", since I don't think it's an analogy at all. It's the same phenomena, as explained below. Any time you have a mechanism for dissipation (a coupling between the "system" and a "bath") that coupling mechanism will give rise to back-action of the bath the ... 5 The phase constant is needed only if you have a specific initial condition, e.g. if I told you where x was at time t = 0, you could solve for \varphi. Otherwise you can just choose whatever you want for it: Note that it is the same in all functions. Choosing some value for \varphi is analogous to you manually setting the time origin to something ... 4 Yes, principally because the speed of sound depends on the temperature. An approximate equation for the speed of sound in dry air is:$$ 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} $$In principle there will be some thermal expansion of the ... 4 The oscillator frequency \omega says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this:$$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$At t=0 it has a phase \phi_0. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force ... 4 No, it is not. Your system will go through the same point twice in every oscillation, once moving in each direction, and the friction force will be reversed in each pass, so your approach doesn't work. What you need to consider is the velocity, not the displacement, so$$ma=-kx - \mathrm{sign}(v) F_{\mathrm{fric}}.$$This is not all that helpful in ... 4 y(\theta) = A\sin \theta+ B \cos \theta is known as the simple harmonic function. All the motions which can be represented by this function are known as simple harmonic motions. Motion of a simple pendulum is approximately a simple harmonic motion for small amplitudes. It stops vibrating after some-time due to drag from air i.e. loss of energy. But, we ... 4 The dimensional analysis in zkf's answer completely solves the exercise. Nevertheless, it is possible to give a closed formula for the period$$ 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

I have just noticed the question. Indeed, the body does have very clear resonances. Nature has prioritised speed of movement over stability so limbs are underdamped and naturally resonant. It is likely that many rhythmic movements occur at the resonant frequency of the body parts involved (rather similar to the oscillation of some insect wings). A ...

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 ...

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