Analyzing the Harmonic Oscillator: Maximal Velocity and Turning Points

[CAF123]

Homework Statement

1)Consider a particle subject to the following force $F = 4/x^2 - 1$ for x>0.
What is the particle's maximal velocity and where is it attained?

2)A particle of unit mass moves along positive x-axis under the force $F=36/x^3 - 9/x^2$
a)Given that E<0 find the turning point(s) as a function of E.
b)Expand the potential about the equilibrium and obtain the period in the harmonic approximation.

The Attempt at a Solution

1) is a part of a bigger question, but only here do I need some advice. I said that the max velocity occurs at E=T => $\frac{m}{2}\dot{x}^2 = E$, so $\dot{x} = \pm \sqrt{2E/m}$. So this must occur when $V(x) = 0 => \frac{4}{x} + x = 0$, which has no solutions. This does not seem right somehow?

2)
a) Just wondering: Why is the plural suggestive? For E>0, the particle is unbounded, but for E<0, then I think there would always exist two points x<x' such that oscillation would occur between these points.
b) So $V(x) \approx V(x_{equil}) + V''(x_{equil})(x-x_{equil})^2/2$. I found equilibrium point x=4 so I think what the question wants is V(x) ≈ -9/8 + k/2 (x-4)². To find the period do I assume T = 2π √(m/k)?

Many thanks

 

[NascentOxygen]

For part (1), have you sketched F? Find the point where F changes from a push to a pull, and that will co-incide with maximum speed.

You probably have a sign wrong in V(x).

 

[voko]

1) is a part of a bigger question, but only here do I need some advice. I said that the max velocity occurs at E=T => $\frac{m}{2}\dot{x}^2 = E$, so $\dot{x} = \pm \sqrt{2E/m}$. So this must occur when $V(x) = 0 => \frac{4}{x} + x = 0$, which has no solutions. This does not seem right somehow?

Total energy is conserved. When velocity is maximal, kinetic energy is maximal. What does that mean w.r.t. potential energy?

a) Just wondering: Why is the plural suggestive? For E>0, the particle is unbounded, but for E<0, then I think there would always exist two points x<x' such that oscillation would occur between these points.

This assumption is baseless. There can be any number of turning points, they don't have to come paired. For example, imagine there is some great impenetrable wall in space and there is nothing else. Then the wall is just one turning point, and motion can continue infinitely in both directions from the wall.

 

[CAF123]

For part (1), have you sketched F? Find the point where F changes from a push to a pull, and that will co-incide with maximum speed.

This occurs at x=2, so then max velocity is $\dot{x} = \pm \sqrt{2/m} = \pm 1$ for m=2.
What was wrong with my way before? I worked out from a previous part that E = 5. So subbing this in does not give me the answer I just got now.

You probably have a sign wrong in V(x).

Looking at my sketch of V(x) I see that V(x) is never zero. So it makes sense that 4/x + x never has any solns. Furthermore, if this is the case, then what I said about at max velocity E=T would be wrong since the particle would always have some potential energy.Yes?

 

[CAF123]

Total energy is conserved. When velocity is maximal, kinetic energy is maximal. What does that mean w.r.t. potential energy?

Then potential energy is the smallest it can be and looking at my graph, this is when V(x)=4. So solving for 4/x + x = 4 => x = 2. Indeed, this is at max velocity so $T + V = E => m/2 v_{max}^2 + 4 = 5$. Solve to get same answer as before. (I think this resolves my problem posted in response to NascentOxygen)

So I think it is right to say that at x=2, potential energy is smallest but the particle has max velocity there.

This assumption is baseless. There can be any number of turning points, they don't have to come paired. For example, imagine there is some great impenetrable wall in space and there is nothing else. Then the wall is just one turning point, and motion can continue infinitely in both directions from the wall.

This is exactly the case here. For small x, the potential is tending to ∞, while for larger x V → 0. But for E < 0, I found two turning points. Is this right? EDIT: (I mean two turning points for each value of E<0)

 

[voko]

This is exactly the case here. For small x, the potential is tending to ∞, while for larger x V → 0.

Not quite so. In my wall example, there is no force (constant potential) everywhere outside the wall, and a very strong repelling force in its immediate vicinity. In case 2), however, there is an attractive force as well, albeit waning with distance. Which is why for certain energy levels you have two turning point for x > 0.

 

[CAF123]

Not quite so. In my wall example, there is no force (constant potential) everywhere outside the wall, and a very strong repelling force in its immediate vicinity. In case 2), however, there is an attractive force as well, albeit waning with distance. Which is why for certain energy levels you have two turning point for x > 0.

Ok, I think maybe what I meant was that in my example when E > 0, the particle is bounded to the left (since V tends to infinity) but not so to the right (since V tends to 0). So here the particle would not undergo oscillations since there is no possible turning point for the particle at larger x. (and hence no confinement)

Are my results good now? What about the period question in 2B)?

 

[voko]

I do not see that you have found the turning points as a function of E.

2B seems good so far.

 

[CAF123]

I do not see that you have found the turning points as a function of E.

Turning points where E = V. So $E = 18/x^2 - 9/x => Ex^2 + 9x -18 = 0$. Solve quadratic to give $$x_{1,2} = \frac{-9 \pm \sqrt{81 + 72E}}{2E}$$

2B seems good so far.

I found k to be 9/32 so T = $2 \pi \sqrt{32/9} \approx 11.8$

 

[CAF123]

I want to try to find this period without using the approximation and show that the results are consistent for small oscillations.
I know $$T = \sqrt{2m} \int_{x_1}^{x_2} \frac{dx}{\sqrt{E - (18/x^2 - 9/x)}},$$ $x_1, x_2$ the two turning points.

Rearranging the part inside the sqrt to $Ex^2 + 9x -18$ I made this equal to $(x_2 - x)(x - x_1)$. So my intergal becomes $$T = \sqrt{2m} \int_{x_1}^{x_2} \frac{ x dx}{\sqrt{(x_2 - x)(x - x_1)}}$$

Now let $x = x_1 cos^2 \theta + x_2 sin^2 \theta$. Differentiating, subbing in and cancelling I end up with two apparently simple looking integrals: $$2 \sqrt{2m} \int_{\theta_1}^{\theta_2} x_1 cos^2 \theta d \theta + 2 \sqrt{2m} \int_{\theta_1}^{\theta_2} x_2 sin^2 \theta d \theta.$$

My question is: How do I solve for the limits?

 

[voko]

Turning points where E = V. So $E = 18/x^2 - 9/x => Ex^2 + 9x -18 = 0$. Solve quadratic to give $$x_{1,2} = \frac{-9 \pm \sqrt{81 + 72E}}{2E}$$

I recommend that you let z = 1/x and solve the original equation for z, and then find x. It will have a more readily understandable dependency on E so you can analyze it further.

I found k to be 9/32 so T = $2 \pi \sqrt{32/9} \approx 11.8$

I have k = 9/64. How did you get yours?

 

[voko]

My question is: How do I solve for the limits?

Obviously, $x = x_1$ corresponds to $\theta = 0$, and $x = x_2$ corresponds to $\theta = \pi / 2$.

 

[CAF123]

I recommend that you let z = 1/x and solve the original equation for z, and then find x. It will have a more readily understandable dependency on E so you can analyze it further.

Ok, I'll try this.

I have k = 9/64. How did you get yours?

Found V''(4), is it what you did?

 

[CAF123]

Obviously, $x = x_1$ corresponds to $\theta = 0$, and $x = x_2$ corresponds to $\theta = \pi / 2$.

I don't see how you got this. Putting x = x1 into the eqn defined for x gives x1 = x2 which is just nonsense. Thanks.

 

[voko]

Found V''(4), is it what you did?

$V''(x) = 108 / x^4 - 18/ x^3 => V''(4) = 108/4^4 - 18/4^3 = 27/4^3 - 18/4^3 = 9/64$

I don't see how you got this. Putting x = x1 into the eqn defined for x gives x1 = x2 which is just nonsense. Thanks.

Let $\theta = 0$. What do you get?

 

[CAF123]

$V''(x) = 108 / x^4 - 18/ x^3 => V''(4) = 108/4^4 - 18/4^3 = 27/4^3 - 18/4^3 = 9/64$

Yes, I made a silly error.

Let $\theta = 0$. What do you get?

Ok, I see. But there is no analytic solution to that eqn (it is transcendental, right?). So you basically solved it by inspection.

Why if I just put x = x1 do I get nonsense?

 

[voko]

Ok, I see. But there is no analytic solution to that eqn (it is transcendental, right?). So you basically solved it by inspection.

Yes, initially I just guessed. But then I made sure the guess made sense. In particular, I computed the derivative and determined that if x2 > x1, the derivative is positive when $0 \le \theta \le \pi/2$, so the function just grows monotonically. That means I can basically map any angular subrange into any [x1, x2] range. I just choose the angles and then solve for the coefficients at the sine and the cosine terms.

If the coefficients are fixed like we have it here, then, due to the monotonicity, there is just one solution. If they are fixed in some other way, we may or may not have a solution, but that requires messy trigonometry to decide.

 

[CAF123]

Thanks voko for your help.
One more question though: Going back to question 1). At E=4,the energy is all potential (since at E=4, this corresponds to the minimum of potential V = 4). Does this mean if a particle had energy E=4 then it would be stationary? And only for E>4 does the particle exhibit oscillatory motion.

 

[voko]

Yes, when the total energy of a system is at a minimum of its potential energy, then the system can only be stationary. This follows directly from conservation of energy.

 

[CAF123]

In the integral method to find the period (non approximation), I get that T = $\sqrt{2 \pi} - \sqrt{2}$. However, the next part of the question asks to verify that this is consistent with what I did in the harmonic approximation.

Can I write that for small oscillations: $cos 2\theta \approx 1 - \frac{(2\theta)^2}{2}$ and $sin 2 \theta \approx 2 \theta$? When I do this, I don't get the solutions to match.

EDIT: Using T = $2\pi \sqrt{m/k}$ gives a period of about 16s which is much larger than what I get when I do the integral properly.

 

[voko]

Can you show your integration steps? I did it earlier today and I am pretty sure I got a different result.

 

[CAF123]

Can you show your integration steps? I did it earlier today and I am pretty sure I got a different result.

Sure, I get to $$2\sqrt{2m} \int_0^{\pi/2} x_a cos^2 \theta d \theta + 2\sqrt{2m} \int_0^{\pi/2} x_b sin^2 \theta d\theta $$
Then evaluate:$$2\sqrt{2m} \left[ \frac{1}{4}sin 2\theta + \frac{1}{2}\theta \right]_0^{\pi/2} + 2\sqrt{2m} \left[ \frac{1}{4} cos 2\theta + \frac{1}{2} \theta \right]_0^{\pi/2}\,\,(1)$$ gives $T = \sqrt{2} \pi - \sqrt{2} = \sqrt{2}(\pi -1)$.

So this is the period in no approximation. In the approximation, T = $2\pi \sqrt{m/k} = 2 \pi \sqrt{64/9} \approx 16$

When I repeat step (1) but with the approximations of sin2θ and cos2θ I gave in my last post, I get about 3 which is consistent with $\sqrt{2} (\pi -1)$ to about 5%. But I don't think I want it to be consistent with this soln - what I want is it to be consistent with the approximation using $2\pi \sqrt{m/k}$

EDIT: If $m \neq 1$ then what I would get is $\sqrt{2m} \pi - \sqrt{2m}$, which is dimensionally incorrect, unless I incorporate some of the units into the numbers. (Which I believe I am allowed to do since the expression for F given in the Q is in fact dimensionally incorrect unless the numbers have some physical dimensions.)

 

[voko]

The integral of cosine squared from 0 to pi/2 is equal to that of sine squared and us equal to pi/4.

And you lost x1 and x2 along the way.

 

[CAF123]

The integral of cosine squared from 0 to pi/2 is equal to that of sine squared and us equal to pi/4.

I get the integral of cosine squared to be pi/4 but I have another term for the integral of sin squared. (pi/4 -1/2)

And you lost x1 and x2 along the way.

Ok, so now I have a period that is dependent on x1 and x2. How am I going to show the consistency if I have no values for these turning points?
EDIT: I see my error for the integral of sin sqaured.

 

[voko]

Recall the expressions for x1 and x2. You all really need to do is add them together.

 

[CAF123]

Recall the expressions for x1 and x2. You all really need to do is add them together.

Yes, I get to $\frac{\pi}{\sqrt{2}} (x1 + x2)$. If $x_{1,2} = \frac{-9 \pm \sqrt{81 + 72E}}{2E}$ when I add x1 and x2 together I will get T = $\frac{\pi}{\sqrt{2}}[-9E]$ which is still a function of E and now negative.

 

[voko]

I think that you should have -9/E in the brackets.

And E must be negative to begin with, otherwise you won't have to positive turning points. Now, what is the the minimum of E?

 

[CAF123]

I think that you should have -9/E in the brackets.

And E must be negative to begin with, otherwise you won't have to positive turning points. Now, what is the the minimum of E?

-9/8?

 

[voko]

Yep.

 

[CAF123]

Yep.

So, I sub this into the period eqn to find T for small oscillations and I get T = 2pi. This is not close to the T=16 I get when I did the approximation with T = $2 \pi \sqrt{m/k}$?

EDIT: It's getting late and I am making so many errors. I get that T = $\frac{8 \pi}{\sqrt{2}} \approx 17$ (not 2pi). So can I say this is about the 16 I get when I do the approximation?

 

[voko]

I do not understand how you get 2pi/ As I said, you should have -9/E in the brackets.

 

[CAF123]

I do not understand how you get 2pi/ As I said, you should have -9/E in the brackets.

See my edit before. It's not that I had -9/E which caused me to get 2pi, it was because I said 8 = 2√2!

 

[voko]

The two values are pretty darn close if you ask me. I am not looking forward to rechecking all the algebra again, anyway :)

 

[CAF123]

Not to actually proceed with this method, but would it be possible to go ahead and use the taylor approximations (sin2θ ≈ 2θ and cos2θ ≈ 1- (2θ)^2/2) at the integration stage to show that for small oscillations I recover the same results?

 

[voko]

I guess the most direct way would be to linearize the original integrand. Your method might work, too.

But, in the end of the day, you got an about 1% difference, which I think is very good.