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1

I think I have since fixed the problem, the issue is the impulse response should be found by the Laplace transform rather than the Fourier transform. Sometimes the two will coincide, but the Fourier transform makes assumptions of periodicity (since we are decomposing the signal into sines/cosines) which are not present for a free particle. If we use a ...


1

You are conflating work with energy, but there is a slight difference. In fact: Work=change in Energy, i.e. U=ΔK. Work is the accumulated force magnitude over distance along a path, so that an element of work dU=<F, ds>=F.ds.cos(θ) , where ds=elemental displacement along trajectory curve S , F=applied Force along S. We integrate this and then call the ...


1

where is the other half of the energy that the battery supplied? Half the energy supplied is dissipated in the resistance that will be present in any real circuit. For a simple RC circuit like below, the switch will be closed at time t=0 and the cap is initially uncharged. The time constant, τ, is RC = 0.05 seconds. So, within 5 time constants (0.25 ...


2

Consider how the capacitor is charged up over time. Naively, since there's no resistance or inductance, the current in the circuit instantly becomes infinite, then instantly shuts off. This is both mathematically ill-defined and unrealistic. To understand what's actually going on, we have to account for nonideal features of the circuit, such as resistance or ...


1

Only your statement "that $F(x)$ is the conservative portion of the force" is incorrect. $F(x)$ is the total force on the mass and thus it must equal $ma$. That's Newton's second law, the sum of the forces on an object is equal to its mass x acceleration. And, yes, you're done (just take out that sentence and put in Newton's second law).


1

This is similar to damped harmonic motion (where the damping is caused by air resistance and friction). The cosine term in your equation represents the oscillatory motion and the exponential part of the equation ("modulates") determines the decay of the amplitude over time. Assuming you have access to one, a sonic motion detector can be used to ...


-1

Just give the pendulum a push, track the angle vs time with a camera or some other time-stamped recording scheme, and then fit the data with your model. Your fit will give you estimates for the various parameters, as well as a confidence interval for each. Is this the kind of detail you’re looking for?


2

No, it is not possible to do, at least not perfectly. Any system complex enough to keep light in a small region is also complex enough to interact thermally with the photons and slowly leak energy to the surrounding environment. Light carries momentum. Imagine a light wave travelling towards the wall of your imaginary box. If the light is to be contained in ...


4

I'll expand quantitatively on Gert's answer. I do this using a greatly simplified example. Let's take a square box of exactly one cubic meter. That way it becomes a matter of how many times the light can bounce inside the box. The best reflectivity of a material I could find is dielectric mirrors. Wikipedia claims it can reach a reflectivity of 99.999% (or ...


6

The speed of light is about $300,000\,\mathrm{km/s}$. A ray of light, even if enclosed in a large box, will bounce off the walls many, many times per second. Now even if we use the most reflective material to coat the box' walls, like silver or platinum, each time the ray hits one of the walls a small amount of the light is absorbed by the wall's coating. ...


2

Initial answer: The second equation is an approximation used when conductivity ($\sigma$) is very small. You can see that in the limit $\sigma\to 0$ the two equations become equivalent. The approximate form is appropriate in vacuum, air, and many other good insulators. Alternate answer: After reading the Wiki page, it appears the Wiki authors actually had a ...


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