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Various types of heat transfer are good examples. A time-dependent temperature field $T(x,y,z,t)$ satisfies $${\partial^2T \over\partial t^2}=\nabla \cdot (\lambda\nabla T),$$ where $\lambda$ is the thermal conductivity, which may vary over space. Or in one dimension, $${\partial^2T \over\partial t^2} = \frac{\partial}{\partial ... 4 You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation. Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The ... 5 From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave,$$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial ...

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Hello handsome poster (why thank you kind stranger). The answer to your question it turns out is in Rovelli's "Quantum Gravity", at least insofar as the free scalar field is concerned. This is done in the following way. As you may recall (from Feynman and Hibbs), through various arguments about doing the path integral as a perturbation on the classical path, ...

1

Perhaps your first question is answered with the following diagram where the displacement of the incident (red) plus the displacement of the reflected (red) must equal the displacement of the transmitted (black)? For your second question the answer is that the vertical acceleration of the rope which is proportional to the vertical component of the tension ...

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If the phase difference between the wave is zero i.e lies in the plane of wave motion the resultant displacement is equal to zero, Thus, $A = 0$, $(L,t)=0$, due to that fact, you can use $$A\sin(kx−\omega t)\to-A\sin(-(kx+\omega t))=A\sin(kx+\omega t)$$ for progressive wave, but nothing can happen when you use cosine rule because give us answer equal to zero ...

4

They are not equivalent. A way to see this is to find the normal modes associated to these solutions. Applying the condition $\xi(L,t)=0$ to $$\xi(x,t)=2 A \mathrm{sin}(kx)\mathrm{cos}(\omega t),$$ you get $kL=n\pi$. This gives $\lambda_n=2L/n$ and $f_n=nv/2L$. Which are the correct wavelength and frequency for the rope with both ends fixed. For the ...

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As long as $\mathbf{B}$ is a continuous (once-differentiable) function, when you look at small enough sizes, $\mathbf{B}$ has a Taylor series, the first term of which is a constant. As you let the loop size shrink, only the constant term matters. But then $\int_S \mathbf{B}\cdot d\mathbf{a}\rightarrow \mathbf{B}\int_S d\mathbf{a}\rightarrow 0$ since the area ...

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The walls do not really matter that much here. I would say that the only points that matter to understand that quote are: The confining potential is conservative so that $K+U = constant$ The confining potential is virtually flat $U = 0$ almost everywhere inside the box (say 99% of the box) except very close to the walls These two conditions are enough to ...

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Here's an image of the 'particle in a box' model: At the endpoints, you can see that there is an infinite potential outside the boundary of the box and zero potential inside. The unrealistic assumption in this model is that every time the particle reaches the boundary, an infinite force repels it to keep it in the box.

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123hoedjevan gives you a wrong answer. The principle of least action states that the physical configuration of the system of fields realizes a minimum of the action with respect to compactly supported variations of the fields which, by the very definition of compactly supportedness, must then vanish on the boundary of the support itself. This in turn means ...

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The real answer is: it doesn't really. Or rather: we can still extract some physics out of it! Lets derive  S[\phi] = \int d^d x \mathcal{L}[\phi,\partial\phi]\\ \delta S[\phi] = \int d^dx \delta \mathcal{L} = \int d^dx \left( \frac{\partial\mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial\mathcal{L}}{\partial \partial_\mu\phi} \delta ...

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The issue is that the underlying classical physics is determined by equations of motion (EOMs) (i.e. Newton's 2nd law), which are common for initial value problems (IVPs) and boundary value problems (BVPs). For BVPs , the EOMs can often alternatively be formulated as Euler-Lagrange (EL) equations of a stationary action principle. The latter approach does ...

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Indeed the problem with boundary conditions, generally speaking, is not well-posed. There are boundary conditions admitting no curves or admitting many curves, satisfying both these conditions and Euler-Lagrange equations. Examples. (1) Think of a particle constrained to stay on a smooth sphere where it can freely move. If you assign the North and the ...

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This is a common misconception about what boundary conditions do and how they do it (for example here). You discussed two types of boundary conditions, Neumann and Dirichlet. In Neumann boundary conditions, we impose that the derivative of the variable normal to the boundary is specified, generally to be zero. With Dirichlet, we impose the value that the ...

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