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5

Hints: OP's exercise is essentially a matter of checking an oscillatory Gaussian integral (3) over the initial position $q_1$. Let $\Delta t_M:=t_2-t_1>0$. To render the Gaussian integral convergent, insert Feynman's $i\epsilon$ prescription $\Delta t_M\to\Delta t_M-i\epsilon$. Or equivalently, Wick-rotate $\Delta t_E:=i\Delta t_M$, where ${\rm ...


3

If you initially give to the bob a velocity $\sqrt{4rg}$, it will actually take an infinite time for the bob to reach the top! A little lesser velocity will cause the bob to stop earlier and come back toward the initial point, while a little greater one will take the bob over the top (the motion will continue, with increasing velocity, to the other side). ...


3

Answer to part 1: This is a common method of solving differential equations, employed by physicists to quickly extract a solution without having to make slow progress using more rigorous methods. The first step is to look for an asymptotic solution (i.e. the limit of large $\xi$ in this case), where the equation is easily solvable. Now, we know that this ...


3

i do not understand the part of clamping the center of the coil or cutting it off The solution method appeals to symmetry. Essentially, the system is symmetric (or even about the center of the spring) and this symmetry is exploited to arrive at an answer. If you have taken a first semester in electrostatics, you may be aware of the method of images. ...


2

Hamiltonian formalism won't help so much because the problem is dissipative. You can solve the homogeneous linear diff. equation with constant coefficients by supposing $x=C e^{\lambda t}$. You will get complex solutions, but you should be able to add them to get a real solution. Now you should have $x(t)$ with two arbitrary constants determined by initial ...


2

They are absolutely connected. In both cases, you have the following components: an "inertial term": this can be mass, or inductance. Something that resists change (in velocity, in current) a "linear force term": the spring ($F=-kx$) or the capacitor ($V=Q/C$). a "displacement term": this is $x$ for mechanical, and $Q$ for electrical. Their derivatives $v$ ...


2

every system that oscillated and that should not osscillated would be an application for damped oscillation. There are many examples in control engineering, but to give you a more daily concurrent example: car shock absorbers. If you drive along a road and hit a chuckhole you don't want your car to jump up and down for half a minute. Since car shock ...


1

Using the sum rule for cosines, we find $$ \rho \cos(\omega t + \phi) = \rho \cos(\phi) \cos(\omega t) - \rho \sin(\phi) sin(\omega t).$$ So we see that $\rho \cos(\omega t + \phi)$ is the same as $A\cos(\omega t) + iB\sin(\omega t)$ when $$ A = \rho \cos(\phi)$$ and $$B = i\rho \sin(\phi).$$


1

"Calculate the trajectory" just means calculate $x(t)$, given the potential energy and the initial conditions.


1

The equation of motion for one mass $m$ attached to a spring with spring constant $k$ (with the other end fixed) on Earth is $$ \ddot{x} + \omega^2 x = g $$ where $2 \pi f = \omega = \sqrt{k / m}$. If it's not obvious that the $g$ term on the right hand side doesn't affect the frequency, notice that if we changed our origin so that $z = 0$ when $x = ...



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