Varying Potential Energy and Amplitude in Unusual Harmonic Motion

[Arkavo]

Homework Statement

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'

Homework Equations

F=-dU/dx
a=F/m

The Attempt at a Solution

none..

 

[darkxponent]

F=-dU/dx
a=F/m

That's a good start. Now form these equations.

 

[rude man]

Just posting to get on the thread.

 

[haruspex]

U=K|X|³
F=-dU/dx

You will need to be careful with the sign when you write out the expression for F as a function of x.

 

[Saitama]

Just posting to get on the thread.

A bit offtopic though. You can use the "Subscribe to this Thread" found under the "Thread Tools" (top right in the OP) menu.

 

[ehild]

Homework Statement

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'

Homework Equations

F=-dU/dx
a=F/m

You may use conservation of energy. The energy can be obtained from the amplitude. Find the velocity v=dx/dt as function of x, and integrate for a quarter period.

ehild

 

[Arkavo]

the equation for v is coming such that it is a cubic in x .. unable to solve it

 

[ehild]

There is no solution in closed form, but the definite integral depends somehow on the amplitude. Show what you got.ehild

 

[rude man]

A bit offtopic though. You can use the "Subscribe to this Thread" found under the "Thread Tools" (top right in the OP) menu.

Thanks. Didn't know.

 

[darkxponent]

@rude man: I see the energy conservation method. But can't it be solved using the equation a= F/m, and then solving the differential equation?

 

[rude man]

@rude man: I see the energy conservation method. But can't it be solved using the equation a= F/m, and then solving the differential equation?

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.

 

[darkxponent]

Sorry, dint see it was ehild. I didn't expect that the equation would come out like this mx'' + 3k/m x^2 = 0. Nor do i know how to solve this equation(never studied any method to solve non linear equations).
Now i an can understand why ehild recommended the energy method :D.

 

[voko]

The energy method should yield an equation of the kind dt =f(x)dx, from which one could get the period, if not explicitly, then as a quadrature.

 

[barryj]

How about a numerical solution? In my attachment, I have (I think) calculated the horizontal acceleration to be..
a = (3Agx^2)/(1+9A^2x^4)
This seems reasonable to me since the horizontal acceleration, to the left or -x direction should be zero when x = inf, and zero when x = 0. So, from a starting point, say (1,1) assuming y = x^3 can we find the time until the acceleration = zero? I haven't done this as yet but it might work.

 

[rude man]

The energy method should yield an equation of the kind dt =f(x)dx, from which one could get the period, if not explicitly, then as a quadrature.

I tried that, starting with mv²/2 + kx³ = kA³, confining to x > 0.

That gives v(x). Then I said dt = dx/v(x) as I think you suggested. Then I said T/2 = ∫[x/v(x)]dx from 0 to A where T = period.

What a disaster that resulted in ...

 

[ehild]

I tried that, starting with mv²/2 + kx³ = kA³, confining to x > 0.

That gives v(x).

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

Then I said dt = dx/v(x) as I think you suggested.
Then I said T/2 = ∫[x/v(x)]dx from 0 to A where T = period.

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

 

[haruspex]

I don't think you need to solve the DE. You just need to find the form of T = f(K, A, m). The function f may include a definite integral which could be calculated in principle. E.g. $\int_0^1\frac{du}{\sqrt(1-u^3)}$
Edit: I see ehild beat me to it

 

[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

Don't know where I picked up that extra x either. Senior moment I guess. And yes, 0 to A is only 1/4 period, not half. Otherwise that's exactly what I did. Thanks.

 

[Arkavo]

maybe we can work on an analogy... A ball (of negligible radius {to cancel rotational energy})that rolls down a slope of equation y=x^3.. though i really don't see how it helps

 

[rude man]

A simple number?

Doing what you suggest you wind up with the integral as A^5/2∫dz/(1 - z³)^1/2 with limits 0 and 1 which is stil a disaster, with elliptic functions, imaginary terms, and singularities for z = 1 (and possibly elsewhere), corresponding to x = A.

Perhaps you could show us the rest ...

EDIT: Ok, with the integral evaluated at z = 0 and z = 1 I see from the plots that we get a real number, and the imaginary part, being constant over that interval, evaluates to zero.

All I can say is, thank God for the graphic displays of wolfram alpha!

 

[rude man]

maybe we can work on an analogy... A ball (of negligible radius {to cancel rotational energy})that rolls down a slope of equation y=x^3.. though i really don't see how it helps

Good analogy, but I join you in not knowing how it helps solve the problem.

Using wolfram alpha and its graphic results, we see that the impossible-looking integral evaluates to a finite real and zero imaginary number. As far as I'm concerned the problem is solved. Do you have any questions about the energy approach ehild suggested way back in post #6?

 

[ehild]

A simple number?

Doing what you suggest you wind up with the integral as A^5/2∫dz/(1 - z³)^1/2

Why is A in the numerator?


\int_0^A{\frac{dx}{\sqrt{A^3-x^3}}}=\int_0^A{\frac{dx}{A^{3/2}\sqrt{1-(x/A))^3}}}=\frac{1}{A^{1/2}}\int_0^1{\frac{dz}{\sqrt{1-z^3}}}
(z=x/A).

According to Wolframalpha, the last integral is about 1.402.

http://www5a.wolframalpha.com/Calculate/MSP/MSP18621f124d5ch6cg4i0c00001c2h70h366557i3g?MSPStoreType=image/gif&s=33&w=258.&h=58 .

ehild

 

[rude man]

Why is A in the numerator?

Another senior moment. It's still in the numerator, as A^-1/2.

\int_0^A{\frac{dx}{\sqrt{A^3-x^3}}}=\int_0^A{\frac{dx}{A^{3/2}\sqrt{1-(x/A))^3}}}=\frac{1}{A^{1/2}}\int_0^1{\frac{dz}{\sqrt{1-z^3}}}
(z=x/A).

According to Wolframalpha, the last integral is about 1.402.

http://www5a.wolframalpha.com/Calculate/MSP/MSP18621f124d5ch6cg4i0c00001c2h70h366557i3g?MSPStoreType=image/gif&s=33&w=258.&h=58 .

ehild

Without something as powerful as wolfram alpha this would not have been solvable, far as I'm concerned. I wonder what the question poser had in mind.

PS how do you do definite integrals in wolfram alpha?

 

[ehild]

Another senior moment. It's still in the numerator, as A^-1/2.



Without something as powerful as wolfram alpha this would not have been solvable, far as I'm concerned. I wonder what the question poser had in mind.

PS how do you do definite integrals in wolfram alpha?

I think the question was how the time period depended on the amplitude. And it is proportional to A^-1/2.

I wrote in the definite integral to Wolframalpha as

integral _0^1 (dz/sqrt(1-z^3))

and that was the result:



ehild

 

[darkxponent]

I think the question was how the time period depended on the amplitude. And it is proportional to A^-1/2.

I wrote in the definite integral to Wolframalpha as

integral _0^1 (dz/sqrt(1-z^3))

and that was the result:



ehild

@ehild: I tried solving the integral on my I own my but just got stuck. Is there a method?

 

[rude man]

@ehild: I tried solving the integral on my I own my but just got stuck. Is there a method?

Take a look at the expression for the indefinite integral in Wolfram Alpha and then decide if you want to try to solve it by yourself! It's a blooming miracle that wolfram can do what it does.

@ehild: thanks again. You did some nice work with the math here all around.

 

[darkxponent]

Take a look at the expression for the indefinite integral in Wolfram Alpha and then decide if you want to try to solve it by yourself! It's a blooming miracle that wolfram can do what it does.

The expression for indefinite integral looks beyond the scope of my knowledge. But I wasn't interested in it either. I just wanted to solve the definite one from(0 to 1). First i thought it can be solved by parts but i couldn't. But the gamma function in the expression given by Wolfram Alpha reminds me of some method used to solve definite integration using beta-gamma function. I am not sure, but i think i have studied earlier(in my first year), that is why i was asking.

 

[rude man]

The expression for indefinite integral looks beyond the scope of my knowledge. But I wasn't interested in it either. I just wanted to solve the definite one from(0 to 1). First i thought it can be solved by parts but i couldn't. But the gamma function in the expression given by Wolfram Alpha reminds me of some method used to solve definite integration using beta-gamma function. I am not sure, but i think i have studied earlier(in my first year), that is why i was asking.

I seem to remeber some definite integrals in my CRC tables and I think a bunch of them involved gamma functions. Oh well, now that ehild has shown us how to do definite integrals via wolfram alpha ... that's not conceptually very different than looking up a definite integral in a table, and a whole lot easier.

 

[darkxponent]

Doing a bit of revision.

http://en.wikipedia.org/wiki/Gamma_function

Just solved using it. Gives the same answer as Wolfrom.

 

[rude man]

Doing a bit of revision.

View attachment 59397

http://en.wikipedia.org/wiki/Gamma_function

Just solved using it. Gives the same answer as Wolfrom.

What substitutions did you make to equate this integral with ∫₀¹dz/(1-z³)^1/2?

I don't see it.

 

[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.