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CM M01 2

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A particle fo mass m moves in a one-dimensional potential $V (x) = - a x 2 + b x 4$ with very light damping. The particle is set in motion with a large initial velocity. Suppose now we measure the period of the motion for each full oscillation, and call these periods $T 1$ , $T 2$ , $T 3$ , $T 4$ , and so on. It is observed that the $T i$ briefly became very large for i near some $i 0$ .

a. Explain what makes the periods get large

b. Obtain a scaling form for $T i$ near $i = i 0$ , valid in the limit of small damping. (A scaling form would be something like $T\sim\log\left|i-i_{0}\right|$ , etc). Hint: Consider first the miton without the friction, $m\ddot{x}=-V^{\prime}(x)$ . Recalling that this motion is necessarily periodic, derive an integral formula relating the period of oscillation to the energy and the turning points $x -$ and $x +$ of the motion.

c. Give an approximate sketch of $T i$ as a function of i.

The potential looks like (in this graph a=b=1):

The periods get large because there is the hump in the middle, coming from the $- b x 2$ term, which dominates for sufficiently small x. Particles with nearly zero energy will have very small velocity here, and so spend a lot of time (since there is only small acceleration as well). This is why the periods become large.

We can write down energy conservation for each period, assuming damping is small:

$E=\frac{1}{2}mv^{2}-ax^{2}+bx^{4}$

solving for the velocity:

$v^{2}=\frac{2}{m}\left(E+ax^{2}-bx^{4}\right)$

So we can find the period by:

$t=\int_{-\delta x}^{\delta x}\frac{\sqrt{m/2}}{\sqrt{E+ax^{2}-bx^{4}}}$

If we take $δ x$ of the integration to be $\ll\sqrt{a/b}$ , we can neglect the $x 4$ term, and get:

$t=\int_{-\delta x}^{\delta x}\frac{\sqrt{m/2}}{\sqrt{E+ax^{2}}}$

which we recognize as the arcsinh:

$t=\sqrt{m/2a}\,\mathrm{arcsinh}\left(\frac{\delta x\sqrt{a}}{\sqrt{E}}\right)$

Considering $δ x$ fixed, when we reduce E we can rewrite the arcsinh:

$\sinh x=y=\frac{e^{x}-e^{-x}}{2}\Rightarrow\ln(2y)=\ln\left(e^{x}-e^{-x}\right)$

For large y, to get positive x, x must be large, which means:

$\ln(2y)\approx\ln\left(e^{x}-1\right)\approx x\Rightarrow\mathrm{arcsinh}\,y\approx\ln(2y)$

Plugging in:

$t=\sqrt{m/2a}\ln\left(\frac{2\delta x\sqrt{a}}{\sqrt{E}}\right)$

Only considering the dependence on E:

$t\sim-\ln\left(E\right)+const$

Now we consider the damping to be linear with rate \kappa :

E i = E 0 - κ i

where $E 0$ is the initial energy. We can write the critical value of i as $i 0 = E 0 / κ$ , so that $E i = κ(i 0 - i)$ , which gives the time dependence: