# Tag Info

## Hot answers tagged harmonic-oscillator

13

Usually, the (sinusoidal) driven harmonic oscillator is damped, and the first two parts of your solution (which depend on the initial conditions, while the third term does not) are transient, i.e. not relevant after a short time. That the solution $$x(t) = \frac{F_0\sin(\omega t)}{m(\omega_0^2 - \omega^2)}$$ cannot be the "full" solution to the equation of ...

4

Clearly the motion of the mass can not be described by a single sine! What's going on here? The general solution to the simple harmonic oscillator is the sum of the unforced response (homogeneous solution) and the forced response. The homogeneous solution is $$x_h(t) = x_h(0) \cos(\omega_0 t) + \frac{\dot x_h(0)}{\omega_0}\sin(\omega_0 t)$$ Thus, ...

3

I know of no such publication. However, this issue may simply be on one hand too trivial and on the other hand too far removed from practical relevance. Let's derive what you are after to see: A creation ladder operator $\hat{a}^\dagger$ for arbitrary states would have to be of the form $$\sum_{n=0}^\infty c_n \left| n+1 \right\rangle \left\langle n ... 3 Repeatedly applying this relation to the ground state is exactly what you need to do. There's nothing more to it. 2 I think The Physics of Musical Instruments (Springer Science & Business Media, 1998) by Fletcher and Rossing would be a good starting point for you. The general physical description of sound rests on the investigation of the impedance changes on the boundaries. For example: the reflection at the end of the string is caused by the discontinuity between ... 2 It's the response of the system to a stimulation at zero frequency. In other words, it tells you the displacement of the system in equilibrium under a time independent force. Let me give an example. Consider a mass on a spring with friction and an external force F_{\text{ext}}(t). The friction force is$$F_{\text{friction}} = -\mu \dot{x}$$so the ... 2 Once the mass is released, the center of mass will move at a constant velocity. Superposed on that is the relative motion of the two masses - first towards each other, then away. They will be in exact antiphase so the center of mass has constant velocity. Your mistake was to set x up as a cosine function - that implies that it is at an extreme of position ... 2 First, let's review the basic ideas of simple harmonic motion (I'm assuming an early university level). Starting with Newton's equation:$$F=ma$$and using Hooke's law$$ma=-kx$$then recognizing that acceleration is the second derivative of position x$$mx''= -kx$$We know that simple harmonic motion is sinusoidal, so we substitute x=\sin(\omega t) ... 2 But$$\frac{\sum_{n=o}^\infty E(n) e^{-\beta E(n)}}{\sum_{n=o}^\infty e^{-\beta E(n)}} \ne \sum_{n=o}^\infty E(n)$$Instead$$\frac{\sum_{n=o}^\infty E(n) e^{-\beta E(n)}}{\sum_{n=o}^\infty e^{-\beta E(n)}} = \frac{\sum_{n=o}^\infty \hbar \omega(n + \frac{1}{2}) e^{-\beta E(n)}}{\sum_{n=o}^\infty e^{-\beta E(n)}}=\hbar \omega\left(\frac{1}{2} + ...

1

It depends whether you're talking about an ideal pendulum or a real pendulum. For an ideal pendulum we can ignore the up and down motion of the bob and consider only sideways motion. In that case the equation of motion is: $$\frac{d^2x}{dt^2} = -\frac{g}{\ell}x \tag{1}$$ where $x$ is the displacement of the pendulum bob, $\ell$ is the length of the cord ...

1

In quantum field theory the vacuum is the vector with no particles. It is defined starting from the Fock space, that is the sum of spaces with any number $n\in \mathbb{N}$ of (identical particles). The space with zero particles is, roughly speaking, one-dimensional, and its basis vector is the vacuum. In quantum mechanics, one can use the Fock space ...

1

Your assumption that the force at A is equal to $mg$ is not correct. It might have been correct if the entire system was in equilibrium. Then you would have been justified in using the principle of equilibrium. However, this is a dynamic system, and you have to apply Newton's 2nd Law instead. To solve this problem, let us define three variables, $x$, $y$ ...

1

No, it can't. The unstretched spring extension is the equilibrium solution to the unforced equation $$m\ddot x+\gamma\dot x=-k(x-x_0),$$ whereas the stretched extension is the equilibrium solution to the forced equation $$m\ddot x+\gamma\dot x=-k(x-x_0)-mg.$$ The equilibrium positions are completely insensitive to the damping constant $\gamma$, since you ...

1

You can understand in a simple way the factor $1/3$ which gives you the approximate solution in the low frequency regime (more on this later) in the following way. Start by writing the kinetic energy of your system as: $$K = \frac{1}{2} m \dot{\delta}(\ell)^2 + \int_0^\ell \frac{1}{2} \lambda \dot{\delta}(u)^2 du$$ where $\delta(u)$ is the displacement of ...

1

The use of the following identities may prove helpful: $$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$ $$\sin(A \pm B) = \sin A \cos B \pm \sin B \cos A$$ For example: $$\cos(\omega t - \pi / 6) = \frac{\sqrt{3}}{2}\cos(\omega t) + \frac{1}{2}\sin(\omega t)$$ Then you have a sum of sines and cosines, which you can use the identities again to turn in ...

1

You can analyze the single pendulum problem using two degrees of freedom(x,y) in cartesian coordinates or you can use one degree of freedom(angle) using spherical(here the same as cylindrical i think) coordinates(normally it is two degrees for the spherical coordinate system as well,but for the single pendulum the radius is constant so it makes our life ...

1

Short explanation: Physical systems are usually dissipative systems. Dissipative systems can be modelled with a 1st order ODE system. If you do your perturbation analysis on this system at the Hopf bifurcation and take higher orders into account, you end up with the Stuart-Landau equation. Hence by derivation, it describes the dynamics of a system ...

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