For what values of total energy is the resulting motion periodic?

[dark_matter_is_neat]

Homework Statement

A point particle of mass m is moving along the x axis and is subject to a force $F(x) = -a e^{-bx} (1 - e^{-bx})$ Where a and b are positive constants with appropriate units.

(a) obtain and sketch the potential V(x) adjusted so that V(x) approaches 0 as x approaches infinity. For what values of total energy is the resulting motion periodic?

(b) What is the period of small oscillations about equilibrium?

Relevant Equations

V(x) = $- \int F(x)$

(a). Getting the potential is straight forward, just take the negative derivative of F(x) and then enforce the condition that the function should approach 0 at x = infinity.
$- \int -a e^{-bx} (1 - e^{-bx})$ = $\int a e^{-bx} - a e^(-2bx)$ = $\frac{a}{b} (\frac{1}{2}e^{-2bx} - e^{-bx}) + C$ Where C is 0 since the other terms in the expression go to 0 at x = infinity. So $V(x) = \frac{a}{b}(\frac{1}{2}e^{-2bx} - e^{-bx})$.
What I'm getting stuck on is figuring out the energy values where the motion is periodic. From my understanding the equation of motion for periodic motion is $\ddot{x} = -kx$ for some positive constant k but the equation of motion for this problem is $\ddot{x} = \frac{a}{m} (e^{-2bx} - e^{-bx})$ which doesn't have some clear condition for it to work out to the form I want. I was thinking of expanding the exponentials in terms of x but then I get stuck with some higher order x terms that I don't want.

(b). This is pretty clear from the equation of motion.
First get the equilibrium value of x which is just when $\ddot{x} = 0$ which happens when $a e^{-2bx} - a e^{-bx} = 0$ which is just true for x = 0, so x = 0 is the equilibrium value. Then you just perturb the equilibrium value by some small value $\epsilon$ and do a taylor expansion of the right side of the equation of motion which results in $m \ddot{\epsilon} = -ab \epsilon + ...$ so the frequency of small oscillations is just $\sqrt{\frac{ab}{m}}$

 

[Orodruin]

What I'm getting stuck on is figuring out the energy values where the motion is periodic. From my understanding the equation of motion for periodic motion is $\ddot{x} = -kx$

No, that is the equation of motion for a harmonic oscillator. A system in periodic motion is much more general than that. It simply means that the system after some period $T$ will return to its initial state.

or some positive constant k but the equation of motion for this problem is $\ddot{x} = \frac{a}{m} (e^{-2bx} - e^{-bx})$ which doesn't have some clear condition for it to work out to the form I want. I was thinking of expanding the exponentials in terms of x but then I get stuck with some higher order x terms that I don't want.

Thus, instead of trying to go this route, you need to figure out a condition that is required for the system to return to its previous state. The typical way of doing this is asking yourself the question: When can the system turn around? (The points where it turns around are known as the classical turning points. How many times does the system need to turn around to return to its original state?

 

[kuruman]

Additional hint
You are given $F(x)$.
You also know that $F(x)=ma$.
Can you find $v(x)$? If so, you can you find the classical turning points that @Orodruin is talking about?

 

[dark_matter_is_neat]

I know that the turning points are going to be at $\dot{x} = 0$ but then I need to solve $\int \frac{a}{m} (e^{-2bx} - e^{-bx}) dt$ to get an expression for the velocity which seems incredibly difficult.

 

[Orodruin]

I know that the turning points are going to be at $\dot{x} = 0$ but then I need to solve $\int \frac{a}{m} (e^{-2bx} - e^{-bx}) dt$ to get an expression for the velocity which seems incredibly difficult.

No, you do not need to solve that to get an expression for velocity. Particularly not for when velocity is zero.

You know what the potential energy is and energy is constant.

 

[haruspex]

I need to solve $\int \frac{a}{m} (e^{-2bx} - e^{-bx}) dt$ to get an expression for the velocity

Only if you want velocity as a function of time, which you do not need.

 

[dark_matter_is_neat]

No, you do not need to solve that to get an expression for velocity. Particularly not for when velocity is zero.

You know what the potential energy is and energy is constant.

So I determine when dV(x)/dt = 0? $\frac{d}{dt}(\frac{a}{b}(\frac{1}{2}e^{-2bx} - e^{-bx}))$ = $a\dot{x}(e^{-bx} - e^{-2bx})$ = 0 which is true when x = 0.

 

[Orodruin]

So I determine when dV(x)/dt = 0? $\frac{d}{dt}(\frac{a}{b}(\frac{1}{2}e^{-2bx} - e^{-bx}))$ = $a\dot{x}(e^{-bx} - e^{-2bx})$ = 0 which is true when x = 0.

No, that is when acceleration is zero. You want to know when velocity is zero. In fact, you do not even need the actual values. You only need to know when there are two classical turning points.

Time derivatives (apart from velocity itself) and actual time dependence is irrelevant.

 

[Orodruin]

So I determine when dV(x)/dt = 0? $\frac{d}{dt}(\frac{a}{b}(\frac{1}{2}e^{-2bx} - e^{-bx}))$ = $a\dot{x}(e^{-bx} - e^{-2bx})$ = 0 which is true when x = 0.

Also, what that tells you is when the time derivative of the potential is zero. This happens either at potential extrema or at the classical turning points $\dot x = 0$ so this doesn’t actually give you anything new.

 

[Orodruin]

So to give one more hint (and anything more would be giving away the solution): Can you write down an expression for the total energy?

 

[Frabjous]

I would suggest plotting the potential.

 

[dark_matter_is_neat]

Looking at the potential plot, I'd assuming that the total energy would have to be less than 0 since the potential approaches 0 from below 0 as x approaches infinity.

Also the expression for energy is just $E = \frac{1}{2}m\dot{x}^{2} + \frac{a}{b}(\frac{1}{2}e^{-2bx} + e^{-bx})$.

Going back to the equation of motion $\ddot{x} = \frac{a}{m}(e^{-2bx} - e^{-bx})$ and multiplying each side by $\dot{x}$ and then integrating with respect to t, I get $\int \ddot{x} \dot{x} dt$ = $\int \dot{x} \frac{a}{m} (e^{-2bx} - e^{-bx}) dt$ use u substitution on each side. On the left side I can use u = $\dot{x}$ so du = $\ddot{x} dt$, so the integral on the left side works out to be $\frac{v^{2}}{2}$. On the right side I can use the substitution u = x so du = $\dot{x} dt$, so the integral on the right works out to be $\frac{a}{mb} (-\frac{1}{2} e^{-2bx}+e^{-bx})$. So $\frac{v^{2}}{2} = \frac{a}{mb} (-\frac{1}{2}e^{-2bx} + e^{-bx})$. Substituting this expression for $v^{2}$ back into the expression for energy I get $E = -\frac{a}{b}(\frac{1}{2}e^{-2bx} - e^{-bx}) + \frac{a}{b}(\frac{1}{2}e^{-2bx} - e^{-bx}) = 0$.

 

[haruspex]

$\frac{v^{2}}{2} = \frac{a}{mb} (-\frac{1}{2}e^{-2bx} + e^{-bx})$.

So where are the possible turning points in the motion? What must also be true there regarding the derivatives of velocity (wrt what?) for them to be turning points?

 

[dark_matter_is_neat]

The turning point is where v = 0 (and $v^{2}$ should also be 0 here). So solving for $v = 0$, $-\frac{1}{2}e^{-2bx}+e^{-bx} = 0$ which occurs when $x = -\frac{ln(2)}{b}$. The time derivative of velocity at this point is non-zero since it's where the velocity changes its direction although it doesn't have to reach some particular value so it's not very useful for determining the total energy required for periodic motion. I don't see any required conditions for the spatial derivative of velocity.

 

[Frabjous]

Looking at the potential plot, I'd assuming that the total energy would have to be less than 0 since the potential approaches 0 from below 0 as x approaches infinity.

Also the expression for energy is just $E = \frac{1}{2}m\dot{x}^{2} + \frac{a}{b}(\frac{1}{2}e^{-2bx} + e^{-bx})$.

Going back to the equation of motion $\ddot{x} = \frac{a}{m}(e^{-2bx} - e^{-bx})$ and multiplying each side by $\dot{x}$ and then integrating with respect to t, I get $\int \ddot{x} \dot{x} dt$ = $\int \dot{x} \frac{a}{m} (e^{-2bx} - e^{-bx}) dt$ use u substitution on each side. On the left side I can use u = $\dot{x}$ so du = $\ddot{x} dt$, so the integral on the left side works out to be $\frac{v^{2}}{2}$. On the right side I can use the substitution u = x so du = $\dot{x} dt$, so the integral on the right works out to be $\frac{a}{mb} (-\frac{1}{2} e^{-2bx}+e^{-bx})$. So $\frac{v^{2}}{2} = \frac{a}{mb} (-\frac{1}{2}e^{-2bx} + e^{-bx})$. Substituting this expression for $v^{2}$ back into the expression for energy I get $E = -\frac{a}{b}(\frac{1}{2}e^{-2bx} - e^{-bx}) + \frac{a}{b}(\frac{1}{2}e^{-2bx} - e^{-bx}) = 0$.

You’ve lost a constant of the integration. Once you add it in, look at the first E equation that you wrote down and think about it.

 

[dark_matter_is_neat]

You’ve lost a constant of the integration. Once you add it in, look at the first E equation that you wrote down and think about it.

Throwing in the constant of integration I get $\frac{v^{2}}{2} = \frac{a}{mb}(\frac{-1}{2}e^{-2bx} + e^{-bx}) + C$ where C is the constant of integration, so substituting this expression back into the expression for energy, I just get that E = Cm, so the energy is constant which is something I already knew and so I just went around in a circle...

 

[kuruman]

. . . and multiplying each side by $\dot x$ and then . . .

Is this wise when at the turning points $\dot x = 0$?
I echo @Frabjous's suggestion to plot the potential. A plot will guide your thinking.

 

[haruspex]

The time derivative of velocity at this point is non-zero

Right, but more generally the condition is that the first nonzero derivative is an odd derivative. E.g. consider $v=t^2$, for which there is no reversal of direction, and $v=t^3$, for which there is.

 

[dark_matter_is_neat]

Is this wise when at the turning points $\dot x = 0$?
I echo @Frabjous's suggestion to plot the potential. A plot will guide your thinking.

The plot of the potential setting all the constants to one starts from some value greater than zero, decays to some minimum negative value (which is at x = 0) and then approaches 0. So looking at the potential plot the particle is not going to turn back around unless E<0, otherwise it will keep going towards x = infinity. So following the plot, my intuition would be that E<0 is the condition for there to be periodic motion.

 

[kuruman]

The plot of the potential setting all the constants to one starts from some value greater than zero, decays to some minimum negative value (which is at x = 0) and then approaches 0. So looking at the potential plot the particle is not going to turn back around unless E<0, otherwise it will keep going towards x = infinity. So following the plot, my intuition would be that E<0 is the condition for there to be periodic motion.

Right. So for E < 0 you get two turning points. That answers part (a). Note that negative total energy means that you have a bound system.

 

[Orodruin]

To be more precise: $-1/2 \leq E < 0$.

The energy is bounded from below by the potential minimum.

 

[haruspex]

To be more precise: $-1/2 \leq E < 0$.

The energy is bounded from below by the potential minimum.

E=-1/2 would, I suppose, be periodic in a degenerate sense. OTOH, we are told the particle "is moving", so E>-1/2 is guaranteed.

 

[Frabjous]

Note that negative total energy means that you have a bound system.

That is by construction. One could always add a constant to the potential without changing the physics.

 

[Orodruin]

OTOH, we are told the particle "is moving", so E>-1/2 is guaranteed.

Not sure about that. I would say it is a matter of interpretation. I read "moving on the x-axis" as referring to a class of motions that are restricted to the x-axis. Physically, the distinction is of course irrelevant.

 

[kuruman]

That is by construction. One could always add a constant to the potential without changing the physics.

Sure, but the instructions in part (a) specify to “sketch the potential V(x) adjusted so that V(x) approaches 0 as x approaches infinity.”

 

[Frabjous]

Sure, but the instructions in part (a) specify to “sketch the potential V(x) adjusted so that V(x) approaches 0 as x approaches infinity.”

Your comment read like it applied in general.