Varying Potential Energy and Amplitude in Unusual Harmonic Motion

[barryj]

In post #19, Arkavo suggested that we use an analogy of a small ball rolling down a slope y = |x^3|.
Let's say we start at x = 1, y = 1. Can the above analysis determine the time for the ball to roll to the bottom of the curve. This should be 1/2 the period of oscillation. Conservation of energy should show that the velocity of the ball at the bottom is 4.429 m/sec.

 

[darkxponent]

putting t = z^3, reduces the expression dz/(1-z^3) to (1/3)*(t^-2/3)*((1-t)^(-1/2))*dt, which is a beta function.

 

[darkxponent]

In post #19, Arkavo suggested that we use an analogy of a small ball rolling down a slope y = |x^3|.
Let's say we start at x = 1, y = 1. Can the above analysis determine the time for the ball to roll to the bottom of the curve. This should be 1/2 the period of oscillation. Conservation of energy should show that the velocity of the ball at the bottom is 4.429 m/sec.

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.

 

[barryj]

This is why I am asking the question. If we assume a ramp y = x^3, the potential energy vs height is the same as in the given problem.since there is no friction, one can determine the velocity at the bottom of the curve at (0,0). The analogy seems good to me. I did a numerical solution as to the time for a mass to start at 1,1 and reach 0,0. I wonder if it is the same as has been calculated above. As a math and science tutor,. I always tell my students to check their work in some manner, like approaching the problem from a different angle. This is what I am looking for here.

 

[barryj]

Oh. BTW, the acceleration function you mention above is almost the same as the one I derived in post #14. Itried a numerical approach then and did not get the correct velocity at the bottom of the curve. In my original post, I only considered the acceleration in the x direction.

 

[barryj]

Oops, the acceleration function in 33 is the same as I derived if A = 1, in the Ax direction anyway.

 

[voko]

The motion of a massive point constrained to move one the curve $y = x^3$ is not similar in any way to this problem. Its potential energy has no minimum, so there is no oscillatory mode.

The analogy with $y = |x|^3$ is even worse, because the velocity vector is not continuous at $x = 0$.

 

[barryj]

Hmmm,.. I must disagree. Look at my attachment. A ball in a trough will oscillate between say (1,1) and (-1,1) passing through (0,0). At (0,0) there is no PE, only KE. At the start, there is no KE, but there is PE = mgx^3. The ball will obviously oscillate. My question is, what is the period of oscillation or the time from (1,1) to 0,0)

 

[voko]

Hmmm,.. I must disagree.

You must not disagree. You must get your things straight. $y = x^3$ does not look like your trough at all.

 

[barryj]

The original problem was.. "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' "

I forgot to put the absolute value on the x^3, and let K = 1 for simplicity then we do have y = |x^3| , yes?

After all the discussion, did the question ever get answered? Is there a relationship between A (or K) and the period?

 

[rude man]

putting t = z^3, reduces the expression dz/(1-z^3) to (1/3)*(t^-2/3)*((1-t)^(-1/2))*dt, which is a beta function.

Got it. Thanks. Nice research!

 

[voko]

I forgot to put the absolute value on the x^3, and let K = 1 for simplicity then we do have y = |x^3| , yes?

At the end of #37 I said why that was a problem, too.

This "analogy" is beyond flawed, drop it.

 

[rude man]

The original problem was.. "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' "

I forgot to put the absolute value on the x^3, and let K = 1 for simplicity then we do have y = |x^3| , yes?

After all the discussion, did the question ever get answered? Is there a relationship between A (or K) and the period?

Yes. Look at posts 18, 22 and 29. The grand synthesis of these is
kT/2m = 1.402/sqrt(A),
T = period and A = amplitude.

If I didn't slip up again.

 

[barryj]

Rude, would you agree or disagree that a ball on a ramp y = |x^3| would apply to this case?
Assuming you agree., then in your equation above, what is m. If it is mass, does it matter

For the case where y = |x^3| if we start with a particle at x = 1, then y = 1. K = 1 . It seems your equation reduces to T = 2 * 1.402. Am I interpreting this correctly?

 

[rude man]

At the end of #37 I said why that was a problem, too.

This "analogy" is beyond flawed, drop it.

I think the analogy is good from x = A to x = 0 at least. For that range,

mv^2/2 + ky = k y0 where k = mg and y0 = kA^3 certainly holds, and with y = kx^3 we get
mv^2/2 + kx^3 = kA^3

which is the basis for the way we solved the problem.

I also don't see why v is discontinuous at x = y = 0 if y = k|x|^3 , though that's a separate concern. How is it discontinuous?

.

 

[voko]

I think the analogy is good from x = A to x = 0 at least. For that range,

mv^2/2 + ky = k y0 where k = mg and y0 = kA^3 certainly holds, and with y = kx^3 we get
mv^2/2 + kx^3 = kA^3

which is the basis for the way we solved the problem.

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

I also don't see why v is discontinuous at x = y = 0 if y = k|x|^3 , though that's a separate concern. How is it discontinuous?.

Find what 'v' is explicitly, you will see.

 

[barryj]

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.

 

[rude man]

Rude, would you agree or disagree that a ball on a ramp y = |x^3| would apply to this case?
Assuming you agree., then in your equation above, what is m. If it is mass, does it matter

For the case where y = |x^3| if we start with a particle at x = 1, then y = 1. K = 1 . It seems your equation reduces to T = 2 * 1.402. Am I interpreting this correctly?

See my previous post # 45. I see nothing wrong with your analogy except that it didn't help to get the answer.

You can't use k=1, why do you? And yes, m matters. My equation does not reduce to what you say - how did you arrive at that conclusion?

The basis for this entire effort is the conservation of energy: mv^2/2 + kx^3 = kA^3 for 0 <= x <= A. The transit time from A to 0 is T/4.

I don't see why it doesn't apply to your curve, but ultimately the integral ∫dζ/(1 + ζ^3)^1/2 has to be evaluated from 0 to 1. ehild used wolfram alpha and darkxponent cleverly recognized the applicability of the beta function to this definite integral. They both got to the answer I posted in post 43.

 

[rude man]

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



Find what 'v' is explicitly, you will see.

That's easy: v = sqrt[(2k/m)(A^3 - x^3)].
As x → 0, v → sqrt(2kA^3/m).

When x goes negative, replace x with |x|. But that's not a necessary consideration for the first quarter-period when x goes from +A to 0.

 

[barryj]

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.

 

[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.