Consider the differential equation

$$\ddot{x}=-n^2 x$$

Now it can be shown that an equivalent formula is


, where $A$ is the amplitude of this simple harmonic motion and $n>0$.

From here, say we want to find $v$, given that the particle starts from the origin and moves to the right at some velocity. The usual technique here would be to make $v$ the subject and to take the positive case, since the particle moves to the right initially.

So we obtain


in other words the velocity is always positive.

However, we know that the particle moves back and forth since the second order ODE is the model for simple harmonic motion.

So say we didn't know that, how would one know that $v$ in fact satisfies both the plus AND the minus case?

Some problems I've done have only satisfied the one case but a naive approach here would have concluded that too.

EDIT for further clarification:

We know that $$\ddot{x}=\frac{d}{dx} \left ( \frac{1}{2}v^2 \right )$$

so if we define $\ddot{x}=f(x)$, then

$$\frac{d}{dx} \left ( \frac{1}{2}v^2 \right ) = f(x)$$

and so

$$v^2 = 2\int f(x)dx$$

$$v=\pm \sqrt{2\int f(x)dx}$$

There are some systems where $v$ is precisely the positive or the negative case (in other words the motion is either purely positive or purely negative), but there some other systems (like simple harmonic motion) where it satisfies both.

How would you determine which is which, without knowing any information about the system prior to doing the question?

  • $\begingroup$ It should be $v=\pm n\sqrt{A^2-x^2}$ not $v=n\sqrt{A^2-x^2}$. $\endgroup$ – user5402 Jun 29 '15 at 12:35
  • $\begingroup$ I'm aware of that. But if the given initial conditions are that it moves to the right initially, then you would take the positive case. $\endgroup$ – Trogdor Jun 29 '15 at 12:36

It should be $v=\pm n\sqrt{A^2-x^2}$ not $v=n\sqrt{A^2-x^2}$. To see that, let $x=A\cos(\omega t)$ then $v=-\omega A\sin(\omega t)$.

At $\displaystyle t=\frac{\pi}{6\omega}$, we have $\displaystyle x=\frac{A\sqrt{3}}{2}$ and $\displaystyle v=-\frac{\omega A}{2}$

At $\displaystyle t=\frac{11\pi}{6\omega}$, we have $\displaystyle x=\frac{A\sqrt{3}}{2}$ and $\displaystyle v=\frac{\omega A}{2}$.

So $v$ is a bifunction of $x$.

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  • $\begingroup$ I have clarified my question further with an edit. I know that $v$ is a bifunction of $x$. My question was moreso aimed towards how we would've known that WITHOUT being familiar with this system being in simple harmonic motion. Surely not all expressions of velocity are bifunctions. $\endgroup$ – Trogdor Jun 29 '15 at 12:54
  • $\begingroup$ That's easy you solve $\displaystyle v=\pm \sqrt{2\int f(x)dx}$, you'll get 2 solutions. You check which one corresponds to the problem. Even better, why not start with $\displaystyle v^2 = 2\int f(x)dx$ and not worry about the $\pm$ signs? $\endgroup$ – user5402 Jun 29 '15 at 13:00
  • $\begingroup$ But that is exactly my question. How do you know which one corresponds to the problem exactly? If the question provides the fact that the particle moves to the right initially, we are awfully tempted to conclude that we just take the positive solution (since it moves positive initially). But my counter example to this logic was the SHM example. How would we have deduced that in the SHM example where concluding that it was the positive case was incorrect, since the system actually takes both cases. $\endgroup$ – Trogdor Jun 29 '15 at 13:03
  • $\begingroup$ As I told you,you solve the 2 cases, you'll get 2 solutions. You check which one corresponds to the problem: $$v=\pm n\sqrt{A^2-x^2}$$ $$t=\int_{A}^{x}\frac{\pm 1}{n\sqrt{A^2-z^2}}dz=\frac{\pm 1}{n}\arctan\left(\frac{x}{\sqrt{A^2-x^2}}\right)\mp\frac{\pi}{2n}$$ $$\frac{x}{\sqrt{A^2-x^2}}=\pm\tan\left(nt\pm\frac{\pi}{2}\right)$$ $$\frac{x^2}{A^2-x^2}=\tan^2\left(nt\pm\frac{\pi}{2}\right)=\cot^2(nt)$$ $$\frac{A^2}{A^2-x^2}=1+\cot^2(nt)=\frac{1}{\sin^2(nt)}$$ $$x^2=A^2\cos^2(nt)$$ Using the initial condition $x(0)=A$, we have $$x=A\cos(nt)$$. $\endgroup$ – user5402 Jun 29 '15 at 13:21

First, observe that the first integral of the equation of motion can be written in the form $(nx)^2+\dot{x}^2=(nA)^2$ with $n,A>0$. This means that the particle's trajectory in the $(x(t),\dot{x}(t))$ plane -- essentially, the phase space of the system--is an ellipse. From here the double-valuedness of $v(x)$ is apparent: there is a trajectory with positive $v$, and another with negative $v$.

Happily, these two halves of the curve are symmetric across the $x$-axis, so it's sufficient to integrate under the assumption that $v>0$ and then obtain the $v<0$ portion by symmetry. Alternatively, we could change from Cartesian coordinates in this space over to polar coordinates, for which only positive $r>0$ is allowed and any ambiguity vanishes. (As an aside, this is the simplest example of what are known as action-angle variables.)

To generalize this, consider a particle subject to some 1D force $f(x)$. Then the argument above shows that the trajectories in phase space are still symmetric across the $x$-axis. But it no longer need be the case that the trajectory in phase space crosses the $x$-axis twice i.e. the particle doesn't have to oscillate). In that case it is entirely possible that only one sign matters.

If you want to see the differences, I suggest playing with the following potentials: $f(x)=x$ (linear potential e.g. gravity on earth), $f(x)=e^{-x^2}$, and $f(x)=e^{-x^4+x^2}$. (The last is more subtle than the first two, since different initial positions lead to different kinds of trajectories.)

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  • $\begingroup$ An aside: The above symmetry argument breaks down for velocity-dependent potentials. But such forces are usually nonconservative and so lack first integrals of motion. (The Lorentz force is an exception to this, but one which requires moving beyond 1D). $\endgroup$ – Semiclassical Jun 29 '15 at 14:52

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