Varying Potential Energy and Amplitude in Unusual Harmonic Motion

[voko]

You did not express the velocity vector in terms of $x$ and $\dot{x}$.

 

[rude man]

rude, I guess I am trying to find a concrete answer to a specific problem. Can you tell me where I am wrong below. I assume a curved ramp, y = |x^3| I start with a ball, no friction, at (1,1)

To easily find the velocity at (0,0) use PE converts to KE.
(1/2)mv^2 = mgh which reduces to

v = sqrt(2gh) = sqrt(2*9.81*1) = sqrt(19.62) = 4.429 m/sec moving in the -x direction.

So, how long did it take to get from (1,1) to (0,0). That is the question.

Well, I agree with your computation but, to quote you, " So, how long did it take to get from (1,1) to (0,0) That is the question ". Your computation only gives the velocity at x = 0. But to compute transit time you need to integrate: ∫dt = ∫dx/v(x) from 0 to A. And that leads to our definite integral as solved by two posters separately.

 

[voko]

And that leads to our definite integral as solved by two posters separately.

It does not. Why are you ignoring the fact that v is different?

 

[barryj]

rude, I totally agree with what you said in #52. Yes, the velocity(speed) at (1,1) = 0 since that is the start. The velocity(speed) at x = 0 = 4.429 m/sec. The time from (1,1) to (0,0) is the question.

What is it?? I am looking for you, or someone to tell me an answer in seconds.

Here is why I want to know. I did a simulation, or propagation starting at (1,1) and continuing to (0,0), My simulation does end with the correct velocity at (0,0) so I am pretty sure I have the correct time. I want to know if the methods used previously involving complicated integrals give the same answer.

 

[rude man]

You did not express the velocity vector in terms of $x$ and $\dot{x}$.

OK, I think I'll give up on the analogy. Obviously, m has to be included in the answer yet if you use the y = kx^3 analogy it does not figure. Thanks.

 

[barryj]

rude, 1) what do you mean that you give up on the analogy.
2) why does m matter when we use conservation of energy to get the velocity?

(1/2)mv^2 = mgh which reduces to

v = sqrt(2gh) = sqrt(2*9.81*1) = sqrt(19.62) = 4.429 m/sec moving in the -x direction.

Somehow, I feel that all of us are picturing totally different views of the problem.

 

[voko]

Not just that. The biggest difference is that $v = \dot{x}\sqrt{1 + 9x^4}$.

 

[rude man]

rude, let's reset a bit. here are to many Ks, ks, As T and etc to keep track of so let's start with as simple case as I can think of namely y = |x^3| and work with this.

I have always learned that conservation of energy means:
Potential energy at the start = kinetic energy at the bottom (neglecting relativistic issues :-) )
this means (1/2)mv^2 = mgh where m = mass, v = speed (not necessarily velocity a vector), g = 9.81 and h = heighth. Using this equation, the velocity at the bottom or anywhere else in the trajectory does not depend on the mass.

Surely, we can agree that a ball (assuming no rotational issues) takes a time to travel from (1,1) to (0,0).
All I am asking is what do you think the time is? in seconds. Is it 1.42 seconds? or what.

I'm afraid I have had to concede that your analogy is inappropriate after all.

By your analogy the transit time is indeed not a function of m, whereas in the OP's problem it definitely is. That was enough for me to reluctantly toss it out.

So your question does not pertain to the OP's problem. But OK, you asked a specific, unrelated question, and the answer is addressed in variational calculus. Specifically, given y = f(x), in a constant-gravity environment, you need to solve

t_A→B = ∫ from x=A to x=B of {[1 + (dy/dx)²]/2gy}^1/2dx.

So for you, A = 1 and B = 0, dy/dx = 3x² and y = x³.

I tried to solve t = ∫ from 1 to 0 of {[1 + 9x⁴]/2gx³}^1/2dx but wolfram says this integral does not converge, which I don't understand at all. Sorry. Maybe later.

 

[barryj]

Here is the OP.
The potential energy of a particle varies as U=K|X|3, it is oscillating and the amplitude is 'A' then find out the time period's variance with 'A'

Please tell me where m fits in? You might think that since "potential energy" is mentioned that mass has to be a factor. However,
The question is how the time varies with the amplitude. Recall that for a pendulum, the period is
2(pi)sqrt(L/g) the mass does not matter. Sure you can talk about the potential energy of the ball at various heights but as far a the period is concerned, mass does not matter and should not be in the final answer.

Guys, picture a physical situation. A ball in a salad bowl. You let it go at the rim and it will oscillate back and forth forever if there was no friction. This is the same thing except the side of the bowl is defined as y = K|x^3|.
It could be defined as y = K|x^10| just as well. Clearly, the problem has a solution. I can build this bowl and measure the time with a stopwatch.

 

[rude man]

Here is the OP.
The potential energy of a particle varies as U=K|X|3, it is oscillating and the amplitude is 'A' then find out the time period's variance with 'A'

Please tell me where m fits in?

Simple conservation of energy.

The particle starts with potential energy P.E.(x=A) = kA^3 but as it moves from x = A to x = 0 it loses potential energy : P.E.(x) = ka^3 - k|x|^3. That means it gains K.E. = 1/2 mv(x)^2. Conservation of energy demands that 1/2 mv^2 + k|x|^3 = kA^3.

You are playing with fire with your analogies which are inappropriate. I know, I got burned myself for a while.

 

[voko]

For any system where kinetic energy is $T = \frac {mv^2}{2}$ and potential energy is some $\Pi(x)$, so that $\Pi(0)$ is a local minimum, there is an oscillatory mode in certain region close to $x = 0$, and, if $\Pi(x)$ even, there is some $A$ the amplitude.

For such systems, the equation of motion is $$ v = \sqrt { \frac 2 m (\Pi(A) - \Pi(x)) } $$

If $v = \dot{x} \sqrt {1 + 9x^4}$ and $\Pi(x) = mg|x|^3$, the equation is $$ \dot{x} \sqrt {1 + 9x^4} = \sqrt { 2g (A^3 - |x|^3) } $$ or $$ \dot{x} \sqrt { \frac {1 + 9x^4} { 2g (A^3 - |x|^3) } } = 1 $$ yielding $$ \int_0^A \sqrt { \frac {1 + 9x^4} { 2g (A^3 - |x|^3) } } dx = \int_0^{T/4} dt = \frac T 4 $$

 

[barryj]

I sort of agree with this. 1/2 mv^2 + k|x|^3 = kA^3 but you have things mixed up. You have m, v, A, and K without any definitions at all. At the start, there is only potential energy , mgh to be exact. At the bottom of the curve, at (0,0) there is NO potential energy. It has been converted to kinetic energy. Between the start and the finish, we have the equation you are stating that (KE at any point) = (PE at the start) - (PE at that point). his is elementary physics.

What I have been saying over and over is that you can calculate the speed (velocity) of the object at the bottom by converting the initial potential energy to kinetic energy. Thus KE at bottom = PE at top.
(1/2)mv^2 = mgh
divide out the m, and solve for v = sqrt(2gh)
If we simplify and specify a specific curve y = K|x^3| where we only consider the first quadrant and let K = 1, we have y = x^3. Now let's assume we start the ball at (1,1) and end at (0,0) . The speed at the bottom is
v = sqrt(2gh) = sqrt(2*9,81*1) = 4.29. I have said nothing at all about the speed or velocity between the start and finish.

The problem here is determining the time from the start to finish and I have a good answer for the time and I want to know what you and the others think the time should be.

 

[barryj]

Is this the longest thread or what :-)
OK, here is the challenge.
I have a ramp defined by height = (distance from the start)^3
I put a ball at a position of distance = 1 meter and of course the height at that point is 1 meter.
I release the ball.
Question 1. What is the speed or velocity at the bottom of the ramp? The answer is 4.429 meters/second
Question 2. How long does it take to get to the bottom of the ramp? The answer if 0.594 seconds.
I challenge anyone to disprove these results!

 

[darkxponent]

@barryj: I would agree with Voko here. This is not an analogy.

Except that 'v' here is not just the time-derivative of x it was previously.

I think this is the reason why this won't work. The motion is in two dimensions and the given problem is just in one dimension.

 

[rude man]

Here is the OP.
The potential energy of a particle varies as U=K|X|3, it is oscillating and the amplitude is 'A' then find out the time period's variance with 'A'

If k = mg then I don't see why your analogy would not work. Then y = k|x|^3, mv^2/2 + mgy = mgy0 where y0 = A|x|^3. So mv^2/2 + mg |x|^3 = mg A^3 and m would drop out of the equation.

But k is not defined that way.

 

[rude man]

Is this the longest thread or what :-)
OK, here is the challenge.
I have a ramp defined by height = (distance from the start)^3
I put a ball at a position of distance = 1 meter and of course the height at that point is 1 meter.
I release the ball.
Question 1. What is the speed or velocity at the bottom of the ramp? The answer is 4.429 meters/second

Question 2. How long does it take to get to the bottom of the ramp? The answer if 0.594 seconds.
I challenge anyone to disprove these results!

barry, if you'll tell us how you arrived at the time of 0.594 s we'd be all ears and eyes.

 

[barryj]

dark, can you describe physically an example of this type of motion?

 

[darkxponent]

Is this the longest thread or what :-)
OK, here is the challenge.
I have a ramp defined by height = (distance from the start)^3

You mean the curve y^2 = (x^2+y^2)^3. I think this would work. But this system can be reduced to one dimensional system(taking 'l' as the only dimension), which would reduce to that of the given question.

 

[barryj]

dark, no the curve is y = x^3 or h = d^3. I assumed that since the OP did not specify a mass, or spring constant, although it might be non-linear, I mentally pictured a ball (ignoring moment of inertia) rolling down a ramp. Clearly the potential energy would be proportional to x^3. You mentioned one dimensional motion. Hmmmm.. I will have to think about that.

rude, I did a simulation with time steps of 0.00001 seconds. I started at point (1,1) and continued until the x and y values were both zero. As a check, I found the speed at the bottom to be 4.429 which agrees with the KE = PE energy relations. Also, the time is slightly greater than if the ball were merely dropped. As I decreased the time step size, starting at T = .1, which is much to course, to 0.00001 sec, the time to the bottom converged to 0.594 seconds. I can give you the code if you like. I wrote it in Visual Basic.

 

[voko]

barry, if you'll tell us how you arrived at the time of 0.594 s we'd be all ears and eyes.

It is given by the integral in #61, if A = 1 and g = 9.8.

 

[barryj]

voko, I tried the integral you cannot integrate from 0 to A=1 as the function is indeterminate at A = 1. In any event I don't know what physically this problem applies to but the answer doesn't make any sense. Try it, give me your answer.

 

[darkxponent]

dark, no the curve is y = x^3 or h = d^3. I assumed that since the OP did not specify a mass, or spring constant, although it might be non-linear, I mentally pictured a ball (ignoring moment of inertia) rolling down a ramp. Clearly the potential energy would be proportional to x^3. You mentioned one dimensional motion. Hmmmm.. I will have to think about that.

The only common thing which you are talking about is the potential energy function is NOT common.

Let 'd' be the distance traveled by the ball.

1-Looking at the given question potential energy is given by U = K*d^3.

2-Potential energy for y = x^3, comes out to be, U = mg*x^3,

where d^2 = x^2 + (kx^3)^2
and d NOT equal to x.

So the analogy won't be for the curve y = kx^3 ,

but i think the analogy would exist for y^2 = (x^2+y^2)^3

as this would give U = mg*d^3, same as that given in question.

 

[barryj]

dark, I guess what you are saying is that the PE is not due to gravity. It seems like it might be like a spring constant where the "PE" of a spring = -kx. I guess we could have a non-linear spring where the PE = -kx^3. This would be in one dimension. In this case, of a spring, then mass would be a factor in determining the period of oscillation.

 

[voko]

voko, I tried the integral you cannot integrate from 0 to A=1 as the function is indeterminate at A = 1. In any event I don't know what physically this problem applies to but the answer doesn't make any sense. Try it, give me your answer.

My answer is given in #61. The numeric answer agrees with yours ±0.001.

 

[barryj]

voko, well I be darned :-) I redid my integration and got the same result.

 

[darkxponent]

dark, I guess what you are saying is that the PE is not due to gravity. It seems like it might be like a spring constant where the "PE" of a spring = -kx. I guess we could have a non-linear spring where the PE = -kx^3. This would be in one dimension. In this case, of a spring, then mass would be a factor in determining the period of oscillation.

I did not say anything about gravity. What i was trying to prove was that the analogy you have given is wrong. Any conservative force having the same function for potential energy(proportional to x^3) for me is identical. I am not bothered about the constant 'k' or 'mg', as they hardly make any difference.

 

[rude man]

It does not. Why are you ignoring the fact that v is different?

I'm slow to wake up. I thought we had the solution but now no one has it unless you think you do.

 

[rude man]

That gives the magnitude of v(x)... |v(x)|=2k/m√(A³-x³)



What is that extra x in the integral? And a quarter period elapses while going from x=0 to x=A.

Integrating v(x), it leads to \int_0^A{\frac{dx}{\sqrt{A^3-x^3}}}=\int_0^{T/4}{\frac{2k}{m}dt}

Factor out A^3/2 from the square root,and introduce the variable z=x/A, then you get an integral which is dimensionless, a simple number (consult Wolframalpha), multiplied by some function of A.

ehild

ehild, when I look at your derivation here it appears you also used dx/dt = v where v is the inertial speed derived from mv^2/2 = kA^3 - kx^3. Voko pointed out much later (post 46) that that's wrong, and I agree it is. So this too looks wrong. Agree?

EDIT: never mind, I forgot that there is only one direction here, namely x.

 

[rude man]

dark, no the curve is y = x^3 or h = d^3. I assumed that since the OP did not specify a mass, or spring constant, although it might be non-linear, I mentally pictured a ball (ignoring moment of inertia) rolling down a ramp. Clearly the potential energy would be proportional to x^3. You mentioned one dimensional motion. Hmmmm.. I will have to think about that.

rude, I did a simulation with time steps of 0.00001 seconds. I started at point (1,1) and continued until the x and y values were both zero. As a check, I found the speed at the bottom to be 4.429 which agrees with the KE = PE energy relations. Also, the time is slightly greater than if the ball were merely dropped. As I decreased the time step size, starting at T = .1, which is much to course, to 0.00001 sec, the time to the bottom converged to 0.594 seconds. I can give you the code if you like. I wrote it in Visual Basic.

OK, that's fine. But bottom line is that no one has the solution to the OP's problem unless voko does or unless ehild still is sticking to his.

 

[voko]

I'm slow to wake up. I thought we had the solution but now no one has it unless you thinl you do.

The solution for the original problem, specifically, the integral, does not solve the new problem. Which can be checked in #61, which has a generic solution and its application to the new problem.

 

[darkxponent]

ehild, when I look at your derivation here it appears you also used dx/dt = v where v is the inertial speed derived from mv^2/2 = kA^3 - kx^3. Voko pointed out much later (post 46) that that's wrong, and I agree it is. So this too looks wrong. Agree?

ehild used v= dx/dt, for the problem asked by Arkavo, and Voko used v = (dx/dt)*sqrt(1+9x^4) for the problem given by barryj. Both Voko and ehild are right!

 

[rude man]

It is given by the integral in #61, if A = 1 and g = 9.8.

Yeah, OK. I'm trying to figure out why my integral had x^3 in the denominator, which integral did not converge, instead of A^3 - x^3 which does.

But we're back to no solution to the OP's problem as far as I'm concerned.

 

[voko]

OK, that's fine. But bottom line is that no one has the solution to the OP's problem unless voko does or unless ehild still is sticking to his.

The only problem I see in ehild's solution to the original problem is that the right-hand side must have the square root of 2k/m.

 

[rude man]

ehild used v= dx/dt, for the problem asked by Arkavo, and Voko used v = (dx/dt)*sqrt(1+9x^4) for the problem given by barryj. Both Voko and ehild are right!

Why is ehold right? His v wcame from kinetic energy conservation which is inertial speed, not dx/dt. But he integrated dt = dx/v, not dx/(dx/dt) = dx/v_x.

 

[rude man]

The only problem I see in ehild's solution to the original problem is that the right-hand side must have the square root of 2k/m.

What about him using inertial v derived from energy conservation instead of dx/dt?

 

[voko]

What about him using inertial v derived from energy conservation instead of dx/dt?

I do not know what you mean. Have a look at the generic solution in #61. ehild's solution (with said correction) can be derived from it, assuming that the body is constrained to move horizontally, and potential energy is $k |x|^3$.

 

[rude man]

I do not know what you mean. Have a look at the generic solution in #61. ehild's solution (with said correction) can be derived from it, assuming that the body is constrained to move horizontally, and potential energy is $k |x|^3$.

I got mixed up between the two problems. The OP's problem can use inertial v because there is only one degree of freedom for the particle, which differs from barryj's case where there are two df.

Sorry for my obstinacy. All is well, at least for the OP's problem. As for barryj's situation, I'm still trying to figure out why my textbook-obtained formula for time did not work out. The difference from your post #61 is I had |x|³ in the denominator and you had (A³ - |x|³). I'll figure it out sooner or later ... thank you for your patience and help.

EDIT: I did figure it out. My textbook |x|^3 was the OP's A^3 - |x|^3. Just a mixup of y axes. All's well.

Signing off on this thing!

 

[voko]

I got mixed up between the two problems. The OP's problem can use inertial v because there is only one degree of freedom for the particle, which differs from barryj's case where there are two df.

Both problems have one degree of freedom. The difference is that in the second problem the body is constrained to move along a curved line, and the curvature translates into additional terms in kinetic energy.

Note that the denominator in the integrand is made solely of potential energy, and so has to be pretty much the same in both problems.

 

[barryj]

Well, I got the answer I wanted even though it might have been for the wrong problem. I keep thinking about how the OP could be implemented and I think a non-linear spring might be an example.Thoughts?

 

[Arkavo]

I don't think this will behave as analogy *** the acceleration function in this situation comes out to be

a = (3gx^2 dx)/sqrt(1 + 9x^4) which is very different from what i get for the given problem.

no it would not acceleration is x" rate of rate of change of x coordinate only you are calculating acceleration along the slope

 

[ehild]

The only problem I see in ehild's solution to the original problem is that the right-hand side must have the square root of 2k/m.

You are rigth, I forgot the square root from the right-hand side. The correct formula is

\int_0^A{\frac{dx}{\sqrt{A^3-x^3}}}=\int_0^{T/4}{\sqrt{\frac{2k}{m}}dt}.

The OP speaks about a particle performing periodic motion with amplitude A, when its potential energy function is k|x|^3. k is a constant. It asks how the time period varies with the amplitude. There is no words about slopes and gravity, but a particle well might have mass.
That potential energy can come from an electric field or the gravity of a specific mass distribution ... anything. What is the sense to consider the problem as rolling down along a slope? The potential energy is a general concept, does not necessarily mean gravitational energy near the Earth surface.

ehild

 

[darkxponent]

no it would not acceleration is x" rate of rate of change of x coordinate only you are calculating acceleration along the slope

I never wrote a=x''. I was trying to say that the net acceleration comes out different from that of OPs problem. What i was trying to prove was that barryj's and OPs problem are not analogy.

That was ehild's post that recommended the energy approach.

I tried what you did, wound up with mx'' + 3k/m x^2 = 0 and could not solve it either. I tried a "guessed" solution of x = Asin(wt) which was disastrous. Evidently, the oscillations are not pure sinusoids, neither should we expect them to be in view of the fact that the ODE is nonlinear.

This diffrential equation can be solved putting a=v*(dv/dx), and it reduces to the same v-x equation of ehild's. So one can use either method. The result is the same.