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Consider an arbitrary potential energy $V(x)$; take $x_0$ to be an equilibrium point, that is, $V'(x_0)=0$. Next, Taylor expand $V(x)$ for $x$ close to $x_0$: $$V(x)\approx V(x_0)+(x-x_0)V'(x_0)+\frac{1}{2}(x-x_0)^2 V''(x_0)$$ The first term is just a constant (ie, irrelevant for energies), and the second one is, by definition, null; therefore we find ...

2

First note the following: If $\psi(x,0)$ is any normalizable wave-function in your Hilbert space, then $\psi(x,t) = e^{-iHt} \psi(x,0)$ (I've set $\hbar = 1$) satisfies the time-dependent Schrodinger equation (we call this "time evolution"). This can be seen by using the fact that the stationary states form a basis for your Hilbert space and so $\psi(x,0)$ ...

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It is a Taylor expansion and it is described here . Your $\alpha$ is really $(x - x_0)$ where $x_0$ is the equilibrium position and the second term is zero due to the fact that we are looking at a minimum of the potential, so the slope (the derivative of $U$) should be zero at the minimum. Higher order terms are ignored because of their small values ...

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It tells you what is the spectral content of the motion. Some examples of when this might be interesting: The object being measured is a point on a guitar string---then it would tell you what note is being played The object being measured is a planet. Then the peak frequency of the motion is the inverse of the planet's year. The object being measured is ...

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For your example of a violin string, you can immediately determine that it is not simple harmonic motion by listening to it. Simple harmonic motion is a pure tone of a single frequency. Violins don't sound like that so you immediately know there are harmonics and it therefore is not a simple harmonic oscillator. As some other people have mentioned, a tuning ...

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Your equation for the damped solution is wrong. In order to match the boundary conditions (initial velocity = 0) you have to add either a phase, or a $\sin$ term. I prefer the phase. If initial velocity is zero, the derivative must be zero: $$A_t = A_0 e^{-t/\tau} \cos(\omega t + \phi)\\ v_t = A_0\left(-\frac{e^{-t/\tau}}{\tau}\cos(\omega t + \phi) - ... 2 You typically have one position and one velocity variable per oscillator. The equation of motion of mass i is m_ix_i''=k_ix_i+coupling terms. If the forces from the coupling terms are small, the frequencies do not change much. If the coupling is large, you will have as many modes as oscillators, but the frequencies can be anything. You wind up finding ... 2 When you have an overdamped oscillator and release the mass from rest with some deflection, it will return to zero without ever crossing (no oscillation). For a lightly damped oscillator, you would see oscillations (zero crossings) whose amplitude becomes smaller with time. Now if your mass is launched towards zero (the equilibrium) with sufficient speed, ... 1 using$$a^{\dagger} |n\rangle = \sqrt{n+1}| n+1 \rangle$$and$$ a |n\rangle = \sqrt{n} |n-1\rangle, $$apply these rules in order:$$ a(a^{\dagger}|n\rangle) = a\sqrt{n+1}|n+1\rangle = \sqrt{n+1}a|n+1\rangle = (\sqrt{n+1})^2 |n\rangle = (n+1)|n\rangle 

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You can use hook's law $F=-kx$ . If $\tau$ is the torque on board , then force at the spring, $F=\frac{\tau}{L}$ and $\tau=I\alpha$ , where $I=\frac{mL^2}{3}$ , M.I of rod about one of its end . U can replace x by $x=L\theta$ and appliying to hook's law yield $I\dfrac{d^2\theta}{dt^2}=-L^2k\theta$ this gives $\omega=\sqrt{\frac{L^2k}{I}}$. Therefore, time ...

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The pure simple harmonic motion is in real life very very rare. There are some cases which are really close (e.g. for engineering purposes). That might be: Small-amplitude oscillation of a mass on a spring (small enough for spring nonlinearities not to be pronounced) or other kinds of these simple or moreless model oscillators. Tuning fork. Strictly ...

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