# Tag Info

4

The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it ...

3

It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ...

2

There is a well-known isomorphism between the linear space ${\mathcal M}_{m, n}$ of $m\times n$ matrices and typical (vectorial) linear spaces ${\mathcal L}_{m\times n}$ of dimension $\text{dim}({\mathcal L}_{m\times n}) = m\times n$. Everything that is valid in ${\mathcal L}_{m\times n}$ has an equivalent in ${\mathcal M}_{m, n}$ and conversely. For this ...

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One always needs to allow the frequency to change, otherwise one gets horrible secular terms. In case of resonances one needs additional tricks. A good mathematical book is ''Perturbation methods in nonlinear systems'' by G.E.O. Giacaglia (Springer 2012). He discusses both the traditional Poincare-Linsted method and more advanced methods based on Lie ...

2

Instead I found (for the logistical equation and the Lorenz system) that the logarithmic plot of the distance between trajectories is constant at the start This should not happen and I cannot confirm this. Here is the separation of two trajectories I get for the logistic map (averaged over 10000 realisations): And here is the same for the Lorenz ...

2

The behavior is approximately constant near the beginning because at very short timescales, the transformation of the phase space induced by the time evolution is basically a deformation where the distances change by a factor of $O(1)$. The middle phase is when the chaos is actually growing and the distances are growing exponentially. At the end, at very ...

1

This is more or less an exercise in chasing definitions. The adiabatic invariant $I$ is defined as $$I\equiv \oint p \frac{\mathrm{d}q}{2\pi}\tag{49.7}$$ where the integral is taken over the path for given $E$ and $\lambda$. The external parameter $\lambda(t)$ is a slowly varying function of time $t$ in $\S49$, but is assumed to be a constant in $\S50$. ...

1

Everything is a function of the angle $\theta$ and its derivatives $\dot{\theta}$ and $\ddot{\theta}$. From there use the chain rule of differentiation. \begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = ... 1 As you noticed, if we use Euler-Lagrange equation on L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy we get\ddot x=0\ddot y = -g$$Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations? The point is that system only has ... 1 I) There are already several good answers. OP is asking about the momentum of the non-relativistic string in the transversal model, which in textbooks usually is given as$$ {\cal L}_T ~:=~\frac{\rho}{2} \dot{\eta}^2 - \frac{\tau}{2} \eta^{\prime 2}. \tag{1}$$II) Let us fix notation: \rho is the 1D mass density; \tau is the string tension; Y is the ... 1 Here's a simple diagram of the Atwood machine you describe. For reference: F_{total} = m_{total} \cdot a Let us call g the gravitational acceleration, m the mass of the less massive "block," and a the acceleration of the "duel-block-system."$$F_{total} = m_{total} \cdot aa = \frac{F_{total}}{m_{total}} = \frac{F_{m} + F_{3m}}{3m + m} = ...

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