Varying Potential Energy and Amplitude in Unusual Harmonic Motion

[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