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The problem:

A point mass m hangs at one end of a vertically hung hooke-like spring of force constant k. The other end of the spring is oscillated up and down according to $z=a\cos(w_1t)$. By treating a as a small quantity, obtain a first-order solution to the motion of m in time. Discuss what happens when $w_1$ is equal to $w_0$.

Attempt at a solution:

I understood the oscillating end of the spring as an oscillation of the equilibrium position of the spring, so I used the following hamiltonian:

$H=1/2m[p^2 + m^2w^2q^2-2m^2w^2qa\cos(w_1t)]$

So obviously, the unperturbed hamiltionian is that of a normal spring. Therefore, the canonical transformation that renders the unperturbed hamiltonian null is:

$q=\sqrt{2\alpha/mw^2}\sin w(t+\beta)$

$p=\sqrt{2m\alpha}\cos w(t+\beta)$

Where $\alpha$ and $\beta$ are the new canonical variables (the first represents the energy of the system and the second represents the phase angle).

Using these transformations gives the form of the perturbed hamiltonian in the new coordinates, and a perturbation of order 1 gives (taking initially $\alpha_0=p_0^2/2m$ and $\beta_0=0$):

$\alpha'=2mw^2p_0a\cos(wt)\cos(w_1t)$

$\beta'=-8ma\sqrt{mw^2\over2\alpha}\sin(wt)\sin(w_1t)$

Which can be integrated directly to give the solutions as a function of time.

When $w_1=w$, one can average the first equation, to get the secular change of the energy:

$\langle\alpha'\rangle=mw^2p_0a$

Which shows that the energy increases by that amount every period.

I did not go any further into the details of the problem, but I am not sure about my solution because if $w_1=nw$, where n is an integer different from 1, then the secular change in energy would be 0, which does not seem reasonable to me. This is my first attempt at a perturbation theory problem. Any comments and discussions are welcome and helpful.

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  • $\begingroup$ Welcome to Physics! Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". $\endgroup$ – Kyle Kanos Jan 9 at 11:02
  • $\begingroup$ It is extremely clear and obvious that i am not attempting to obtain a solution to a "worked problem". I am capable, as clearly demonstrated, of obtaining a solution by myself. I noted in the last paragraph that i find my solution weird, because for certain frequencies, a driven oscillator will in fact not gain any energy. I was hoping for someone with more knowledge than me to comment about that peculiarity, but instead all I get is people who don't understand the problem telling me that i shouldn't post such posts. $\endgroup$ – Celeritas Jan 12 at 14:28

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