Varying Potential Energy and Amplitude in Unusual Harmonic Motion

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The discussion revolves around determining the time period of oscillation for a particle with potential energy described by U=K|X|^3. Participants explore various approaches, including using force and acceleration equations, conservation of energy, and numerical methods. The consensus is that the time period is dependent on the amplitude and can be expressed as T = f(K, A, m), with some integrals leading to complex forms that require numerical solutions. The energy conservation method is highlighted as a viable approach, yielding a relationship between the time period and amplitude, specifically indicating that the period is proportional to A^(-1/2). The conversation emphasizes the challenges of solving nonlinear differential equations and the utility of computational tools like Wolfram Alpha for evaluating integrals.
Arkavo
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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..
 
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Arkavo said:
F=-dU/dx
a=F/m

That's a good start. Now form these equations.
 
Just posting to get on the thread.
 
Arkavo said:
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.
 
rude man said:
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.
 
Arkavo said:

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
 
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the equation for v is coming such that it is a cubic in x .. unable to solve it
 
There is no solution in closed form, but the definite integral depends somehow on the amplitude. Show what you got.ehild
 
Pranav-Arora said:
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.
 
  • #10
@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?
 
  • #11
darkxponent said:
@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.
 
  • #12
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.
 
  • #13
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.
 
  • #14
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.
 

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  • #15
voko said:
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 ...
 
  • #16
rude man said:
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)

rude man said:
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 A3/2 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
 
  • #17
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
 
  • #18
ehild said:
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 A3/2 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.
 
  • #19
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
 
  • #20
A simple number?

Doing what you suggest you wind up with the integral as A5/2∫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!
 
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  • #21
Arkavo said:
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?
 
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  • #22
rude man said:
A simple number?

Doing what you suggest you wind up with the integral as A5/2∫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
 
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  • #23
ehild said:
Why is A in the numerator?
Another senior moment. It's still in the numerator, as A-1/2. :smile:

\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?
 
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  • #24
rude man said:
Another senior moment. It's still in the numerator, as A-1/2. :smile:



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
 
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  • #25
ehild said:
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?
 
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  • #26
darkxponent said:
@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.
 
  • #27
rude man said:
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.
 
  • #28
darkxponent said:
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.
 

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