Let's assume an $xy$ plane and let there be a force field defined by the potential $$V=F_0|x|$$ Though the potential is not differentiable still its a perfectly realisable system. If we solve the force equation with the initial conditions $x = \delta$ and $\dot{x}=0$, we will have to solve it for $x\geq0$ and $x\leq 0$ separately and whenever the particle crosses $x=0$ we will have to switch solutions. Rather than going through that pain is there any approximation that can reduce it to simple harmonic? (since we already know a small perturbation would lead to oscillatory behaviour)
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This
we already know a small perturbation would lead to oscillatory behaviour
is not enough. Harmonic oscillations aren't just oscillatory; they also have a period which is independent of the initial amplitude. Your system doesn't satisfy this, so it can't be understood as a harmonic oscillation.
For the system you propose, if you release the particle at rest from a separation $s$, it will 'fall' parabolically to the origin in time $t=\sqrt{x_0/a}$ (where $a=F_0/m$ for a potential of the form $V(x)=F_0|x|$), and the motion will just be time-translated and reflected copies of this parabolic motion, which will therefore have a period $$ T=\frac{4}{\sqrt{F_0/m}}\sqrt{x_0} $$ that depends critically on the oscillation amplitude $x_0$. This is inconsistent with harmonic motion.
Now, is there some change to the motion that you could do so that it will approximate as harmonic? Sure, there's plenty of similar potentials, like, say $$ V(x) = F_0d \sqrt{1 + (x/d)^2}, $$ which looks white similar to your potential as long as $d\ll x_0$, but then the harmonic approximation is only valid in the limit $x_0 < d$. Or, put it another way, anything that restores harmonic motion would destroy the key aspects of your potential's behaviour.
answered Dec 12, 2017 at 19:11
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2$\begingroup$ Fun fact: For a particle moving in a 1D potential proportional to $|x|^n$, the period obeys the proportionality $T \propto x_0^{1-n/2}$. It's easy to see from this relationship that the $n = 2$ case is special, since it's the only case for which $T$ is independent of $x_0$. $\endgroup$ Commented Dec 12, 2017 at 19:38
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$\begingroup$ Is there an easy way to see the relationship $T \propto x_{0}^{1-(n/2)}$ ?? @MichaelSeifert $\endgroup$ Commented Dec 13, 2017 at 3:08
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$\begingroup$ @PrinceMS: Not an easy one, no. The method I know involves using energy conservation to write $dx/dt$ as a function of $x$; this is then a separable ODE. You can then use a technique like Emilio's to write an equation for the quarter-period in terms of an integral over $x$, from 0 to $x_0$. Clever redefinitions of variables then allow you to pull all the dimensionful parameters outside the integral, allowing you to infer the above-mentioned proportionality. $\endgroup$ Commented Dec 13, 2017 at 4:04
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2$\begingroup$ @MichaelSeifert You don't need to go to energy and quadratures - you can just do the scaling transformation directly on the equation of motion. Or, put another way, if $V(x)=C|x|^n$, then $\ddot x=(C/m)|x|^n$ and dimensional analysis tells you that $[C/m]=[FL/L^nM]=[L^{2-n}/T^2]$, i.e. the EOM constant has nonzero dimension of length unless $n=2$. $\endgroup$ Commented Dec 13, 2017 at 9:50
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1$\begingroup$ @PrinceMS In other words, you can scale the length up by some factor $\lambda$ and the period up by $\lambda^\frac{2-n}{2}$, and the dynamics will not change (which you can also see from the EOM directly). That is then sufficient to show that the period is proportional to the $(1-n/2)$th power of the amplitude. $\endgroup$ Commented Dec 13, 2017 at 9:50