How to find the frequency of small oscillation of a particle under gravity that moves along curve $y = a x^4$ where $y$ is vertical height and $(a>0)$ is constant?

I tried comparing $V(x) = \frac 1 2 V''(0) x^2 + \mathscr O(x^3) = \frac 1 2 kx^2$ (assuming $V(0)$ is ground state and $V'(0)$(slope) remains is horizontal at extrememum, but unfortunately $V''(0) = 0$. I am pretty much clueless. Thanks for your help.

ADDED:: Letting $m=1$ the Lagrangian is $L = \frac 1 2 (\dot x^2 + \dot y^2) - gax^4 = \frac 1 2 (\dot x^2 + (4 ax^3 \dot x)^2)-gax^4 = \frac 1 2 \dot x^2(1 + 16a^2x^6)-gax^4$

The above Lagrangian gives the equation of motion as $$\ddot x(1+16a^2x^6)+\dot x^2 96 a^2x^5 + 4 gax^3 = 0$$

Since we are considering the system a small oscillation whose potential is of order 4, $\mathscr{O}(x^{k>4})$ can be ignored which reduces into $\ddot x = -4agx^3$.

To solve this, $$\frac 1 2 \frac{d}{dt}(\dot x^2) = -\frac{d}{dt}(agx^4)$$ which gives $\dot x = \sqrt{k - 2agx^4}$, Assuming the system begins from $t=0$ at $x=x_0$ with $\dot x = 0$, $k = 2agx_0$, which turns the integral into $$\int_0^{T/4} dt= \int_0^{x_0} \frac 1 {\sqrt{2agx_0^4-2agx^4}} dx=\frac{1}{\sqrt{2ag}x_0}\int_0^1\frac{1}{\sqrt{1-y^4}}dy$$

Which gives $$T = 2 \sqrt 2 \sqrt{\frac{\pi}{ag}}\cdot \frac{\Gamma(5/4)}{\Gamma(3/4)}\cdot \frac 1 {x_0}$$

which is a dubious result. Please someone verify it. Any other methods are welcome.

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    $\begingroup$ Note that a particle moving in one dimension under the influence of a quartic potential $V(x)$ is different than a particle constrained to move along the graph of a quartic function under the influence of gravity. $\endgroup$ – joshphysics Aug 27 '13 at 8:11
  • $\begingroup$ @joshphysics can't I express $V$ as function of $x$ since height is linear function of height and height is the quartic function of $x$? $\endgroup$ – hasExams Aug 27 '13 at 8:20
  • $\begingroup$ Are you using Lagrangians or newton's laws? $\endgroup$ – Spaderdabomb Aug 27 '13 at 10:13
  • $\begingroup$ @Spaderdabomb I am using Lagrangian $\endgroup$ – hasExams Aug 27 '13 at 14:21
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    $\begingroup$ You went wrong by a factor of four in in $d(4 a g x^4)/dx$. It should just be $d(a g x^4)/dx$. And then you forgot a factor of $\sqrt{2}$ when you got $\dot{x}$ by itself - remember the $1/2$ kicking around on the left hand side! Other than the arithmetic mistakes your method is sound. :) You should get the same final answer with the 4 replaced by $2\sqrt{2}$. $\endgroup$ – Michael Brown Aug 29 '13 at 3:51

You got the correct answer on your own, but I'd like to point out a useful treatment for this type of questions, from Landau and Lifschitz, Course of theoretical physics.

It will get you only the result up to a numerical constant, but it's beautiful and absolutely general.

I quickly recall the definition of an Euler homogeneous function: $f(x)$ is said to be homogeneous of degree $p$ if $f(\alpha x) = \alpha^p f(x)$.

Absolute generality

Let $L = K - U$ be a lagrangian. Usually, the kinetic energy is quadratic in the velocity, so that it is homogeneous of degree 2 in space and -2 in time. Or, more clearly, if you "stretch" the time by a factor of $\alpha$ and the space by a factor of $\beta$ the kinetic energy will transform as $$ K^\prime =\frac{\beta^2}{\alpha^2}K. $$ Now, if the potential energy is homogeneous of degree $p$ in space you will get similarly $U^\prime = \beta^p U$, so that the lagrangian becomes $$ L^\prime = \frac{\beta^2}{\alpha^2}K - \beta^p U = \frac{\beta^2}{\alpha^2}(K - \beta^{p - 2}\alpha^2 U). $$ That is equivalent to the original lagrangian whenver $\beta^{p - 2}\alpha^2 = 1$, or $$ \alpha = \beta^{1 - p/2}. $$ When two lagrangians are equivalent, any solution of one is also a solution of the other. So, in this large class of problems when you stretch the distances by a factor of $\beta$, you have to multiply the time by a factor of $\beta^{1 - p/2}$. If you have a period of time $T$ for a solution, then it must be proportional to the distance with the above dependence $x^{1-p/2}$.


The current question

You derived the approximate lagrangian with a quadratic kinetic energy and a potential proportional to the fourth power of the distance from the origin. Hence $p = 4$ in the above formula, and you get straight away $T \propto 1/x$. With dimensional analysis you can actually get $$T \propto \dfrac{1}{\sqrt{ag}x}.$$

Isochronous oscillations

For a potential proportional to the square of the distance (small oscillations of a pendulum, spring), $p = 2$, and $T$ doesn't depend on the amplitude of the oscillations.

Kepler's third law

Gravity has a potential with $p = -1$. Hence $T \propto x^{3/2}$.


The motion in such a curve is quite hard to calculate, and even more so if you do not want to get into the messy details of Jacobi elliptic functions like $\text{sn}(u|k)$. However, for the case of small oscillations there is a simple scaling argument that lets you calculate the dependence of the period on the amplitude.

Consider, then, the case of small oscillations, which you correctly calculate by neglecting the effect of the vertical kinetic term $\tfrac12 m\dot y^2$, and which gives the differential equation $$\ddot x=-4 {ag} x^3.\tag{1}$$ A few words on this equation. For one, it can be explicitly solved in terms of the Jacobi elliptic function $\text{sn}$, which is actually quite harmless. It behaves quite differently to a harmonic oscillator, as can be seen from the phase-space portraits in the Wolfram Alpha link - they would be ellipses in the harmonic case, and are much more 'square' here. Most importantly, because it is anharmonic, its solutions will not be isoperiodic, meaning that the period will depend on the amplitude $x_0$.

To find this dependence, you simply need to re-scale time and position: change $x$ for $\lambda x$ and $t$ for $\mu t$, and your equation will transform into $$\frac{\lambda}{\mu^2}\ddot x=-4\lambda^3 {ag}x^3,$$ and this has the exact same form as (1) if you set $\mu=1/\lambda$. What this means is that if you have some solution with $x(0)=x_0$ (and $\dot x(0)=0$ for simplicity) with period $T_0$, then you can scale-transform it to match the solution to any other initial condition $x(0)=x_1$ (and $\dot x(0)=0$, of course), with scaling $\lambda=x_1/x_0$, with the caveat that you must also transform the time to $\mu t=\frac{x_0}{x_1}t$, which in the bottom line means the new period is $T_1=\mu T_0=\frac{x_0}{x_1}T_0$. Slightly more elegantly, it means $$\text{the product }x_0T_0\text{ is constant for all solutions.}$$ Get the solution for one $x_0$, and you have the numerical constants, but most importantly, you know the crucial thing, which is the scaling: $$\boxed{T\propto {1}/{x_0}.}$$


To elaborate on Michael Brown's comment, you can indeed combine this with some dimensional analysis to get the period's dependence on $a$ and $g$. The period $T$ can only depend on $a,g$ and $x_0$, which means that it must be a function of the characteristic time $\sqrt{x_0/g}$ and the dimensionless quantity $ax_0^3$. Coupled with the scaling above, you are forced into the relation $$T=\frac{C}{\sqrt{ag}x_0},$$ where $C$ is now a dimensionless quantity, probably of order 1.

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    $\begingroup$ +1 Elegant. I noticed that dimensional analysis doesn't completely solve the problem because the dimensionless quantity $ax_0^3$ but I then didn't think to try scaling. Once you know the $1/x_0$ scaling you can get the rest of the parametric dependence on dimensional grounds. $\endgroup$ – Michael Brown Aug 30 '13 at 15:33

Is it fine if I solve it using Newton's laws? You can then maybe convert it to a Lagrangian. I will assume the path on which it oscillates is almost flat for small oscillations

At a displacement of $x$($\lim_{x\to0}$), the gravitational force on it will be $$F_{mg}=-mg\sin\theta$$ where $$\tan\theta=\frac{dy}{dx}=4ax^3$$ As $x\to0$ , $\tan\theta\to0$ so we can approximate $\tan\theta\approx\sin\theta$

Thus we have $$\ddot x=-4agx^3$$

You get a differential equation which is solvable(probably a lot easily using lagrangian equations!)

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    $\begingroup$ Mathematica knows how to solve it in terms of JacobiSN functions (whatever those are!). A more useful method is to use conservation of energy to get a first order d.e. which can be solved by quadrature. All systems which can be reduced to 1D can be solved this way. $\endgroup$ – Michael Brown Aug 28 '13 at 13:21
  • $\begingroup$ Thanks for the hint @MichaelBrown, will keep that in mind for future problems. Right now, I haven't yet learnt to solve differential equations(Except the common Simple Harmonic Motion differential equation) in high school. $\endgroup$ – udiboy1209 Aug 28 '13 at 15:58
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    $\begingroup$ @udiboy: if you want to cheat and show that your above equation is the same thing, multiply both sides by $\dot x$ and integrate $\endgroup$ – Jerry Schirmer Aug 28 '13 at 16:53

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