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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 ...

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As $T = t_1 - t_0 \to 0$, we have$$\lim_{t_1 - t_0 \to 0} \langle x_1, t_1\,|\,x_0, t_0\rangle = \lim_{T \to 0}\left({{m\omega}\over{2\pi i\sin\omega T}}\right)^{1/2}\text{exp}\left[{{im\omega}\over{2\sin\omega T}}\left(\left(x_1^2 + x_0^2\right)\cos \omega T - 2x_0x_1\right)\right]$$=\lim_{\epsilon = iT/m \to 0} ... 3 Beginning with \begin{align} -\frac{\partial H}{\partial x} & = -kx = \dot{p}\\ \frac{\partial H}{\partial p} & = \frac{p}{m} = \dot{x} \end{align} One takes the time derivative of both sides of the second equation \frac{d}{dt}(\frac{p}{m}) = \frac{d}{dt}\dot{x} giving \frac{\dot{p}}{m} = \ddot{x} Substituting from Hamilton's first ... 3 A string is a one-dimensional object, and not a zero-dimensional like a point. This means that the object traced out by it following its trajectory is a two-dimensional surface, the worldsheet, while particles - points - trace out one-dimensional objects, the worldlines. Much of the difference between string theory and theories with particles as the ... 2 The propagator does satisfy the Schrodinger equation for most values of x and t... The easiest way to show this is to let \hbar=m=\omega=1. Also, we can work with \sqrt{2\pi}e^{i\pi/4}K\to K instead of K to clean up the mess a little further. Further, let x_f\to x and x_0 \to 0. Then let: K=fe^{ig} with f=\frac{1}{\sqrt{\sin(t)}} and ... 2 In your comment you asked a second (closely related) question about a cart accelerating horizontally. Both these problems can be solved by transforming the frame of reference to a new frame in which the net acceleration is vertically down. The transformation you want to use is a rotation of the axes; the new "apparent gravity" is the vector sum of the two ... 2 OP asks in the title (v1) for help understanding what the Hamiltonian signifies for the action compared with the Euler-Lagrange equations for the Lagrangian. It seems relevant in this context to point out that there is an action principle for both the Lagrangian and Hamiltonian formalism. On one hand, the stationary action principle for the Lagrangian ... 2 I know I'm late to the party, but let me show that Lagrangian and Hamiltonian mechanics are compatible by direct substitution of the Lagrangian into Hamilton's equations. The Hamiltonian is, in terms of the Lagrangian,H(p,q)=p\dot q(p,q)-L(q,\dot q(p,q))$$Now we take the first Hamilton equation and plug in$$\dot q=\frac{\partial H}{\partial p}=\dot ...

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The energy of a photon doesn't change when moving from one medium to another as pointed out by Andrew in a comment. Considering that $E = h\nu$, $\nu$ being the frequency of the photon and $h$ Planck's constant, we see that the frequency has to stay the same when going from one medium to another. Since the frequency is the same, then the wavelength of the ...

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The net force acting on the mass M will be $$F=-2kx\sin\phi$$ Here,x is extension of spring,which can be found out by $$d=(x+d)\cos\phi$$ $$2d\sin^2\frac{\phi}2=x\cos\phi$$ $$2d\cdot (\frac {\phi}2)^2=x$$...Considering very small oscillations. Also, $$F=M\frac{d^2((x+d)\sin\phi)}{dt^2}=M\cdot d\cdot\frac{d^2(\tan\phi)}{dt^2}$$ If,we consider the ...

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Since this is a 1D problem the frequency of small oscillations is simply given by the relation $$\omega^2 = \frac1m U''(x_0),$$ where $U(x)$ is the potential energy of the system and $x_0$ the equilibrium position. This is explicitly given by $$U(x) = mgx + k\left(\sqrt{x^2+d^2}-d\right)^2,$$ and from here one finds $$\omega^2 = \frac{2k}m\left(1-\frac dl + ... 1 The damping introduces a dissipative element to the system, that is, energy is leaving the spring-mass system with time. As such, the energy is not conserved. The maximum energy for the system occurs at its initial configuration; the spring is stretched to some extent A (initial amplitude). The initial energy for the spring is thus purely of a potential ... 1 I randomly had this typed up in personal notes. Was probably an exercise somewhere. Consider a harmonic oscillator, which is described by the Hamiltonian$$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2q^2$$Doing the Legendre transform, we obtain the action as$$\mathcal{S}=\tfrac{1}{2}m\int_0^t(\dot{q}^2-\omega^2q^2)dt'$$Now we use the Euler-Lagrange equation to ... 1 Because if the coupling is not there it still will behave like$$ m \frac{{\rm d}^2x}{{\rm d}t^2} + m \omega_0^2 x = 0 The coupling just acts as an additional effect of $k (2 x)$ magnitude.

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So this is a bit tricky actually and perhaps the original question makes this more clear. The problem I'm having when thinking about this is what exactly does the function generator do. Does it take the original force of the string and add some force $F(t)$ on top of it, or does it essentially fix the force on one side of the spring? So if the spring is ...

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Think of the function generator producing prescribed displacements $X(t)$. What is the extension of the first spring? It is $x$, so its restoring force is $-k_1 x$. Now what is the extension of the second spring? It is $X(t)-x$, with its restoring force $-k_2 (X(t)-x)$. What is missing from the diagram is the direction of positive forces and displacements. ...

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