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Just consider what happens to a classical simple harmonic oscillator. The object moves fast in the middle, goes to the outermost position, stops there, then goes back. Since it stops at the outermost position, it's much more likely to be found near that position. I.e. if we were to take a photo of the oscillator, we'd most likely find the object in the ...

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What makes a standing wave is having waves in a single medium propagating in opposite directions at exactly the same speed, and in phase. This requires the waves to possess a unique speed through the medium in order to establish the "standing wave" condition (and it also requires the medium to have a finite length into which an integer number of ...

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As you've said, $A$ and $B$ are not necessarily real, they can be complex. In fact, if you want a physical solution, you don't need the $\sin$ term to be $0$, but rather you want to impose $\overline{x(t)} = x(t)$, which is equivalent to $\overline{A} = B$. From here, you can rewrite the last part of the solution as: $$x(t) = e^{-\gamma t} 2\, \mathrm{Re}(A)\... 2 So when this happens, the equations are telling you that there is something physical that you are not considering. You can't prove this particular statement because this particular statement is not true in all physical circumstances. So that should lead you to look at the equations again and see in what limit they might be true, in what limit they might be ... 2 The general solution for your ODE is:$$x(t)=(a+i\,b)\,e^{-\gamma t}e^{i\omega_d t}+(a-i\,b)e^{-\gamma t}e^{-i\omega_d t}\tag 1$$Where a=x(0) and b=D(x)(0) are the initial conditions expand equation (1) you obtain:$$x(t)=2\,e^{-\gamma t}(a\,\cos(\omega_d t)-b\,\sin(\omega_d t))\tag 2$$thus the solution is real you can also write the solution equation (... 2 Part 1 is simply a consequence of the geometry of the scenario, and has nothing to do with its dynamics. If you draw a line from one mass (at, say, A) to the centre of the circle at O then OAP is an isosceles triangle with side \frac D 2, \frac D 2, l and acute angle \frac \alpha 2. It follows that \displaystyle \cos \left( \frac \alpha 2 \right) =... 2 The phrase "An instantaneous force imparts momentum p_0 to the system." is a bit ambiguous, the entire concept of a force is not well-defined in quantum mechanics in general. So what the phrase exactly means is somewhat open to interpretation, and your interpretation differs from the one by the author of the solutions. Before the impulse happened,... 2 Simply harmonic motion occurs when the restoring force is proportional to the displacement from equilibrium. Mathematically, you get this whenever you have a differential equation of the form$$\ddot x=-\omega^2 x$$where \omega^2 is some constant (this is your a=-\omega^2 x). There are many physical systems whose equation of motion has this form: masses ... 2 The period of oscillation of a mass-spring system is given by:$$\boxed{T=2\pi \sqrt{\frac{m}{k}}}$$So the weight of the bob m\mathbf{g} has no effect on it, only its actual mass m. It would be different if it was a swinging pendulum. To answer the central question the OP asked, we need to solve the Newtonian equation of motion. Armed with that ... 1 Gravity pulls stretching the spring down to a new equilibrium position. This equilibrium stretch gives an upward force which balances and cancels the gravity force. The mass when disturbed from the new equilibrium always pushes back toward equilibrium. The Hooke proportionality is still the spring constant k assuming the spring didn’t get stretched too much ... 1 In circular motion we user the Greek letter \omega (omega) to represent angular velocity, so the angle \theta travelled through at time t is \theta = \omega t Typically \omega is in radians per second, so the time taken to complete one rotation is \displaystyle t_{rot} = \frac {2 \pi} {\omega} and the reciprocal of this is the frequency of ... 1 what is the need to use the same variable for angular frequency as that of angular velocity that we use in different concepts? Honestly as long as you know what you are talking about, you can take up any variable as per your convenience. It does not matter. If it is bothering you so much, then use english letter "W" for angular velocity and Greek ... 1 In other to answer your question, i will let you know that simple harmonic motion follows a particular definition (which i believe you know). When ever we study systems or we wish to model a system and it conforms to our real definition of simple harmonic motion, we immediately use the equations that relate. For a mass spring system, we observe its motion, ... 1 For the first doubt, there is a problem in the very same chapter in the initial section which asks you to prove the statement. It all boils down to normalization of the solution. As a general statement, we want our \hat{H} to be positive definite so that our system is unitary, it's used in QFT as well. For the second one since our potential goes to ... 1 The standard way to model the internal energy dissipation (i.e. hysteretic damping, or in viscoelastic materials) for oscillatory motion is to make the stiffness term a complex number. The imaginary part of the number represents the energy dissipation. The energy dissipation depends on the amplitude of vibration, but unlike viscous damping, if the amplitude ... 1 I have this answer for you: Point I: I use the \pi/2 angle those:$$D\,\cos(\alpha/2)=l$$Point II: The total Potential Energy is:$$U=-M\,g\,l\cos(\varphi+\alpha/2)\tag 1$$obtain the Taylor series for \varphi\mapsto \Delta\varphi$$U\approx M\,g\,l\,\left(-\cos \left( 1/2\,\alpha \right) +\sin \left( 1/2\,\alpha \right) \Delta \varphi +1/2\,\cos \...

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Actually, in the standing waves theory this expression will add almost nothing to your study, it becomes really significant only when you deal with traveling waves. As @niels nielsen said, if you want, you can see too the standing waves as a superposition of traveling waves, but, as you can see, even seeing from this side requires the idea of ​​waves ...

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You need to find the solutions for $t> t'$ and for $t> t'$ and then impose the integration constants by the boundary conditions. The problem is actually very similar to that of a delta-function barrier in quantum mechanics, but with zero energy. The solution with sines and cosines does not belong here, since it comes from the theory of ordinary ...

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The differential equation for SHM in 1D is given by \begin{equation} \ddot{y}+\omega^2y=0 \end{equation} Its solution can be written as \begin{equation} y(t)=A\sin(\omega t+\phi) \end{equation} where $A$ and $\phi$ is determined by the initial conditions (say $\phi=0$, which means that oscillator starts from equilibrium position). Then, \begin{equation} y(t)=...

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