Spring with oscillating support (Goldstein problem 11.2)

[Celeritas]

Homework Statement

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 $w_0$.

Homework Equations

The first order approximations to the time variations of $(\alpha,\beta)$ are given by:

$\alpha'=\frac{\partial \delta H}{\partial \beta}$
$\beta'=\frac{\partial \delta H}{\partial \beta}$

Where the result is evaluated at the unperturbed values.

The Attempt at a Solution

[/B]
I understood the oscillating end of the spring as an oscillation of the equilibrium position of the spring, so I used the following hamiltonian, where a term involving $a^2$ has been removed:

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

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 $\beta_0=0$):

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

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

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.

 

[haruspex]

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.

I'm no expert on this, but I would have expected it to increase without bound for odd n but cancel out to zero for even n.
Maybe you could post the details at how you arrived at that.

 

[Celeritas]

I'm no expert on this, but I would have expected it to increase without bound for odd n but cancel out to zero for even n.
Maybe you could post the details at how you arrived at that.

Replacing $w_1=nw$ in the equation of $\alpha'$, and then averaging over the period T of the natural oscillator gives:

$<\alpha'>=2mw^2p_oa\frac 1 T\int_{0}^T cos(nwt)cos(wt)\sim\delta_{n1}$

By orthogonality of the cosine function over the interval [0,T].
P.s., i could not find the proportionality symbol in the LaTeX guide, so i just used $\sim$

 

[haruspex]

Replacing $w_1=nw$ in the equation of $\alpha'$, and then averaging over the period T of the natural oscillator gives:

$<\alpha'>=2mw^2p_oa\frac 1 T\int_{0}^T cos(nwt)cos(wt)\sim\delta_{n1}$

By orthogonality of the cosine function over the interval [0,T].
P.s., i could not find the proportionality symbol in the LaTeX guide, so i just used $\sim$

Ok, that looks right,
Did you try \propto: $\propto$?

 

[Celeritas]

Ok, that looks right,
Did you try \propto: $\propto$?

No I didn't, but that's what I was looking for! thanks!

 

[TSny]

... 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).

In the above equations, $\beta$ would have units of time rather than a phase angle. Maybe you meant to move $w$ inside the parentheses.

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 $\beta_0=0$):

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

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

The coefficients of the trig functions on the right sides don't appear to have the correct dimensions if $\alpha$ is energy and $\beta$ is an angle.

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.

This seems OK to me. The system is an undamped, driven harmonic oscillator. If you consider the solution when $w_1=w$ (i.e., resonance), the energy will continue to build up over time. The graph below shows this for a particular choice of parameters such that the natural frequency of the oscillator is f = 1 Hz and it is driven at 1 Hz.

If the same oscillator is driven at 2 Hz, then the energy varies with time as

You can see that the energy returns to its initial value after each unit of time (i.e., each multiple of the natural period of the oscillator). This turns out to be true for any driving frequency that is an integer multiple of the natural frequency (other than resonance). Since you averaged $\alpha '$ over one period, it is maybe not surprising that you get zero for this average when n is a positive integer different from 1.

[Edit: figures have been updated to correct an error in calculating the energy. The specific shapes of the graphs depend on the choice of initial values for position and velocity.]

 

[Celeritas]

In the above equations, $\beta$ would have units of time rather than a phase angle. Maybe you meant to move $w$ inside the parentheses.

The coefficients of the trig functions on the right sides don't appear to have the correct dimensions if $\alpha$ is energy and $\beta$ is an angle.This seems OK to me. The system is an undamped, driven harmonic oscillator. If you consider the solution when $w_1=w$ (i.e., resonance), the energy will continue to build up over time. The graph below shows this for a particular choice of parameters such that the natural frequency of the oscillator is f = 2πω = 1 Hz and it is driven at 1 Hz.

https://www.physicsforums.com/attachments/236996If the same oscillator is driven at 2 Hz, then the energy varies with time as

https://www.physicsforums.com/attachments/236997

You can see that the energy returns to its initial value after each unit of time (i.e., each multiple of the natural period of the oscillator). This turns out to be true for any driving frequency that is an integer multiple of the natural frequency (other than resonance). Since you averaged $\alpha '$ over one period, it is maybe not surprising that you get zero for this average when n is a positive integer different from 1.

You're correct, my bad, I've made a couple of mistakes. Firstly, $\beta$ isn't the phase angle itself, but it's related to it by $$\tan{\omega\beta}=m\omega\frac {q_o}{p_o}$$

so obviously, it has units of time.

Also, in the $\alpha'$ and $\beta'$ equations, there is a factor of $\frac 1m$ missing, adding that factor corrects the units.

Finally, i guess my solution is correct! it's very interesting how driving an oscillator in that particular way doesn't really change the energy in the long run.

Thank you!

 

[TSny]

Yes, it is interesting. As long as you are not exactly on resonance, the amplitude of the motion will remain finite. So, the energy in the system remains bounded.

For the case of resonance, for certain initial conditions, the energy of the system can actually decrease at first and then increase without bound as shown below.

[Edit: Figure corrected]