# 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.

31

Part of it is that 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 calculus ...

23

When you pluck the string you excite many many overtones, not just the fundamental. You can observe this by suppressing the fundamental. Pluck the string while holding a finger lightly at the center of the string. That point is an antinode for the fundamental and all odd harmonics, but a node for the even harmonics. Putting your finger at that point ...

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 ...

21

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 ...

15

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) ...

10

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.

9

I think Crawford 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 French. Also, David Morin has a set of drafted chapters of ...

8

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 ...

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 $$\frac{d}{dt^2} \left[ \begin{array}{cc} x_1 \\ x_2 \end{array} \right] =\left[ \begin{array}{cc} - 2 \omega_0^2 & \omega_0^2 \\ \omega_0^2 & - \omega_0^2 \end{array} \right] \left[ \begin{array}{cc} x_1 \\ x_2 ... 7 This is an answer by an experimentalist who had been fitting data with mathematical models since 1968. When fitting data one goes to the simplest mathematical models. When the data display variations in time and space the Fourier expansion is extremely useful because it gives the frequencies and amplitudes that will fit a periodic data set. One gets as ... 7 It would depend on damping effects being taken into account or not. Invoking Newton's 2nd Law of motion, a differential equation for the motion of a damped harmonic oscillator can be written (including an external, sinusoidal driving force term): m\frac{d^2x}{dt^2}+2m\xi\omega_0\frac{dx}{dt}+m\omega_0^2x=F_0\sin\left(\omega t\right) Where m is the ... 6 I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian L=L(q,\dot{q}) of n generalized coordinates$$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$One of the first questions one would like to ask is, if a specific ... 6 Actually, it does behave exactly like a pendulum. The equations of motion are exactly the same. The issue, as Jasper pointed out, is damping. When you think of a "normal" pendulum, you are considering a lightly damped oscillator. The balloon, as I will prove below, is heavily damped. For a damped harmonic oscillator (which a pendulum approaches for small ... 6 From here, how do I define the "resonance"? At resonance, the energy flow from the driving source is unidirectional, i.e., the system absorbs power over the entire cycle. When \Omega = \omega_0, we have$$\phi(t) = \frac{A}{2\beta \omega_0}\sin\omega_0 t$$thus$$\dot \phi(t) = \frac{A}{2\beta}\cos\omega_0 t$$The power P per unit mass ... 6 Maybe the pendulum would be an interesting example. The equation of motion of an ideal pendulum of length \ell in a uniform gravity field is$${d^2\theta\over dt^2}+{g\over \ell} \sin\theta=0$$It is well known that in the small angle approximation this reduces to a harmonic oscillator, but the precise solution involves elliptic integrals, and can be ... 6 In refraction and reflection the incoming electromagnetic wave causes the electron density of the refracting material to oscillate. This happens because at any point in space the wave produces an oscillating electric field (and magnetic field, though that isn't relevant here) so any material that has a non-zero polarisability will respond by developing an ... 6 For most systems, if you are operating near equilibrium you are at a point where the net force is zero. That means that for small displacements, there will be a small force proportional to the displacement which restores the system to its equilibrium position (Taylor expansion - for small displacements, only first order effects matter).$$F(x + dx) = F(x) + ...

6

Mathematical demonstration It's straightforward to see why this happens if you use a bit of linear response theory. Consider a generic damped harmonic oscillator. There are three forces, the restoring force $F_\text{restoring} = - k x(t)$, the friction force $F_\text{friction} = - \mu \dot{x}(t)$, and the driving force $F_\text{drive}(t)$. Newton's law says ...

6

Just put some weight on a very sensitive scale before the car passes by. When the car passes by, note the maximum reading on the scale. If there is any upward force exerted, then the maximum scale change will give you the maximum upward force per (base area of the scale).

5

Here is a solution for a spring force that varies directly with displacement. It thus varies with time implicitly, but has no explicit dependence on time or any other variable. Givens and Assumptions oscillator with mass $m$ amplitude of oscillation $A$ oscillator displacement, $x$, varies with time, but $x(t)$ is unknown spring applies force varying with ...

5

Suppose we take two identical pendulums. The green one is undamped and the red one is damped: The force $F_g$ on the green pendulum bob is (approximately) given by the usual simple harmonic oscillator law: $$F_g = -kx$$ where $x$ is the displacement, and the acceleration is just $F_g/m$. Now consider the red pendulum bob. This is damped, so the force ...

5

This question reminds me of some remarks in pages 14-16 of Peter Woit's quantum mechanics and representation theory notes Basically, imagine you are looking at all periodic functions from the real to the complex numbers. This is equivalent to looking at all functions from the unit circle to the complex numbers. Let's add the property that we want our ...

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 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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