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## Hot answers tagged oscillators

14

An exponent cannot have a unit. It is by definition (unless you redefine it for a particular context) the number of repeated multiplications of the base number by itself. It doesn't make much sense to multiply $e$ by itself "1 metre times" or something like that. It must be a unitless number. If it isn't, then you are missing a correction factor. ...

7

As the post you've linked says, any argument of functions such as $\cos$ and $\exp$ must be dimensionless. That is to say, $bt/m$ is dimensionless, as are $\omega t$ and $\varphi$. These parts of the function cannot affect the units of the result. The dimension of $x(t)$ comes from $A_0$, which has dimensions of 'angle' (which is really dimensionless).

5

Assume the total energy of the particle is $E$, a conservative quantity: $$E = \frac{1}{2}m v^2 + A|x|^n.$$ We can obtain the velocity as function of position: $$v = \pm \sqrt{\frac{2(E-A|x|^n)}{m}}$$ Fot the periodic motion, the particle will run back and forth in between two turning points, $$x_r = \pm \left( \frac{E}{A}\right)^{1/n}.$$ The ...

4

The function $\exp(x)$ is usually define as dimensionless and it is define only for dimensionless number $x$. In your case, $−b\cdot t/2m$ is dimensionless and so is $e^{−b\cdot t/2m}$. Similarly, $ωt+φ$ and $\cos(ωt+φ)$ are dimensionless. In conulsion, you get that the units of $x(t)$ are just the same units as the units of $A_0$.

4

This is Kapitza's pendulum. It is not surprising that the pendulum is all over the place when one pumps the system with an external force. The surprise is instead that the inverted position is a stable equilibrium for the effective potential of the pendulum. For details, see the Wikipedia page.

4

In short, pressure. For an acoustic wave (a wave traveling through air), this figure will most commonly refer to the pressure. But that's mostly because pressure is the thing that's easiest to measure in acoustic waves. The velocity of the particles also follows the same graph so it could be a graph of velocity of the particles as well. (That is, the ...

3

Now what I'm not able to comprehend is what exactly this wave represents. Does it represent the oscillation of a single particle in a medium? Or does it represent the wave traveling through the medium as a whole? The sinusoidal curve itself can represent both; however, in this particular example, it represents the shape of the wave throughout the medium as ...

3

Typically, when we look at a travelling sine wave, as your diagram shows, we think of particles moving up and down, this is what is called transversal motion. It is transversal because the oscillation of the particles is perpendicular, that is transversal, to the direction of travel. However, sound waves are not transversal waves, they are longitudinal waves ...

3

The trick is the units of $b$. The requirement that the input to the exponential is dimensionless implies that, together with the form of the exponent, $$[b] = [MT^{-1}]$$ since $[\frac{t}{m}] = [TM^{-1}]$. So the damping constant has units of mass flow rate (e.g. kilograms or grams per second), and that cancels the units of $t$ and $m$ and leaves a ...

3

A pendulum doesn't use exactly the same quantities as the link you posted. Don't use an equation without examining its context and limitations! Examining the traditional damped HO differential equation for a simple pendulum with small angle approximation: $$m r^2\frac{\mathrm{d}^2 x}{\mathrm{ dt}^2}+\frac{b}{2}\frac{\mathrm{d}x}{\mathrm{dt}}+mgrx=0$$ we see ...

2

I think that you can obtain analytical solution for the period $~T$ follow the solution from @ytlu with A and m equal 1 and omit the abs() function the energy E is: $$E=\frac{1}{2}\,v^2+x^n$$ with: $$E_0(v=0~,x=x_0)=E(v~,x)$$ you obtain $$v=\sqrt{2(\,x_0^n-x^n)}$$ thus for $~v=0~,x_T=x_0$ and the period T is: $$\frac{dx}{dt}=v(x)$$ $\Rightarrow$ $$T=4\,\... 2 x and F can each be expressed as a Fourier integral:$$x(t)=\int x(\omega)e^{i\omega t} d\omegaF(t)=\int F(\omega)e^{i\omega t} d\omega$$This of course assumes that F(t) is square integrable (L^2 Hilbert space). The rest is just substitution into the equations of motion and comparison of coefficients (which is possible due to the ... 2 For what it's worth, here is a rough estimate for n\gg 1:$$\begin{align} \frac{|x_n|-|x_{n+1}|}{|x_n|^3} ~\approx~&\frac{x_n^2-x_{n+1}^2}{2x_n^4}\cr ~=~&\frac{E_n-E_{n+1}}{2E^2_n}\cr ~\approx~&-\frac{\langle \dot{E}\rangle}{E^2_{n+1/2}} \frac{T}{4}\cr ~\sim~&-\frac{\dot{E}_{n+1/2}}{E^2_{n+1/2}} \frac{T}{8}\cr ~=~&\frac{2\dot{x}^4_{n+1/...

1

The dot means "derivative w.r.t. time". Thus, we use the chain rule as well as the abbreviation $t^\prime=t-s$ and write $$A \;x(t^\prime) = A \;x(t-s) = \dot x(t-s) = \frac{dx(t-s)}{dt} = \frac{d(t-s)}{dt} \cdot \frac{dx(t^\prime)}{dt^\prime}\Big|_{t^\prime = t-s} = \frac{dx(t^\prime)}{dt^\prime}\Big|_{t^\prime = t-s}$$ and for $y$ and $z$ ...

1

What you have shown in the picture is not a wave but rather a state of deformation of a sound carrying medium (i.e. a description of what you have referred to as the "whole medium") at a specific instant in time. Just like for Newtonian point mass dynamics you also need to specify initial velocities (deformation velocities) in addition to initial ...

1

The maximum charge on the capacitor depends on the current and the frequency. If the frequency goes up, the current will have less time on each cycle to charge the capacitor. This suggests the the maximum charge may occur at a frequency less than the resonant frequency. (The rate of change of the current is small near the peak of the resonance curve.) To ...

1

Generally the coupling (mutual inductance) is maximized if the coils are coaxial. This is related to the fact that, for a given separation, the field along the axis of a small coil or magnet is twice the field on its equatorial plane. The dipole-dipole interaction (which relates to small coils with relatively large separation) gives some insight into how ...

1

To maximize the power transfer between two coils you want to maximizes the faction of flux from coil 1 that passes through coil 2 (parallel to the axis of symmetry). The coils might be coaxial or they might be wound on a loop of ferromagnetic material. If one coil were perpendicular (and centered on the other), the flux linkage would be zero.

1

First consider the idea of normal mode in classical mechanics: it is pattern of oscillation in which every part of a many-body system oscillates at the same frequency. Now take a natural generalisation to a continuous field, and in the case of the electromagnetic field you have a distribution of amplitude where the whole pattern is oscillating at a single ...

1

Yes. The term $b$ is in the position of $\omega_0^2$. And you need to convert $$\sin^2(2 t) = \sin^2\left(\frac{4t}{2}\right) = \frac{1-\cos 4t}{2}$$ to observe the driving frequency.

1

My result is equivalent to that from JalfredP, but I considered just a segment of the hoop measured to an angle $θ_m$ on either side of the top position. I'm looking for the angular frequency of oscillation: $ω^2$ = mgL/I where L is the distance from the pivot point to the center of gravity, and I is the rotational inertia about the pivot point. For the ...

1

Work done of a conservative vector field (such as gravity) force, depends only on body path endpoints, it can be stated mathematically as : $$\oint _{C}\mathbf {F} \cdot d{\mathbf {r} }=0.$$ I.e. path integral over closed loop is zero, so total work done is zero. Thus only swing heights matters, it has nothing to do with angles or exact swing path how it ...

1

Not sure if this will help, but motivated by your note that $\dot x^3$ is small, lets introduce a second time scale $\tau = \epsilon t$ and rescale the variable $x = \epsilon^\alpha y$. Plugging this in $$y_{tt} + 2 \epsilon y_{t\tau} + \epsilon^2 y_{\tau \tau} + y = -\epsilon^{2\alpha} (y_t + \epsilon y_\tau)^3.$$ We seek an expansion in \$y = y^0 + \...

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