# Oscillation of a particle on a parabolic surface [equation of motion]

happyparticle
Homework Statement:
equation of motion
Relevant Equations:
F = ma
##\ddot{\theta} + \frac{k}{m} x = 0##
Hi,

I have a particle on a parabolic surface $$y = Ax^2$$ and I have to show that the frequency is $$\omega = \sqrt{2Ag}$$

I don't know how to deal with a parabola. I don't think I can use the polar coordinates like a circle.

I don't see how to start this problem and in which coordinates system.

For a circle I can use the polar coordinates and then use ##F = ma => -mg sin \theta = mR\ddot{\omega}##

I have to get the equation of motion $$\ddot{\theta} + 2Ag \theta = 0$$

thus, ##2Ag = \omega^2##

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which coordinates system.
It’s straightforward enough in Cartesian.
Draw a diagram of the particle at some (x,y) and analyse the forces.

happyparticle
I think using energy ##u(y) = mgy## can get the job done. Someone pointed out to me. I didn't notice.

However, is it the only way?

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I think using energy ##u(y) = mgy## can get the job done.

happyparticle
##y_{max}, E = mgy_{max}##

##K = E - U, \frac{1}{2}m\dot{x}^2 = mgy_{max} - mgy##

##\dot{x} = \sqrt{2g(y_{max} - Ax^2)}##

I'm stuck here, Am I in the right path?

I'm wondering if ##y_{max} = Ax_{max}^2##

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##\frac{1}{2}m\dot{x}^2 = mgy_{max} - mgy##

happyparticle
##\dot{y} = 2A\dot{x}##

##\dot{x} = \frac{\sqrt{2g}}{2A}##

That doesn't work and I can find why.

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##\dot{y} = 2A\dot{x}##

##\dot{x} = \frac{\sqrt{2g}}{2A}##

That doesn't work and I can find why.
No, I mean as part of the KE.

happyparticle
No, I mean as part of the KE.
That's what I did.

##2A\dot{x} =\sqrt{2g} (y_{max} - y)##

##\dot{x} = \frac{\sqrt{2g}}{2A}(y_{max} - y)##

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##\dot{y} = 2A\dot{x}##
That equation is wrong. You have not differentiated x2 correctly. Use the chain rule.
##\frac{1}{2}m\dot{x}^2 = mgy_{max} - mgy##
That equation is wrong because it omits the KE embodied in the vertical component of motion.

happyparticle
If ## \frac{1}{2}m\dot{y}^2 = mgy_{max} - mgy##

I get ##x = \frac{\sqrt{2g}}{2A}(y_{max} - y)##

or ##\dot{y} = \sqrt{2g}(y_{max} - y)##

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If ## \frac{1}{2}m\dot{y}^2 = mgy_{max} - mgy##
Now you are only counting the vertical component of KE. Both contribute.

happyparticle
Now you are only counting the vertical component of KE. Both contribute.
Ah, I see, but how I do that.
##K = \frac{1}{2}m\dot{y}^2 + \frac{1}{2}m\dot{x}^2##

##U = mgy##

##E = mgy_{max} + mgx_{max}##

Which gives me

##\dot{y}^2 + \dot{x}^2 = 2g(y_{max} + x_{max} - y)##

I'm not sure what to do with the left side.

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##E = mgy_{max} + myx_{max}##
I assume you meant mgxmax, but I don't understand the reason for that term. What aspect of GPE is not covered by mgy?

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happyparticle
E = total energy

When the particle is at ##y_{max}## then all the energy = U, thus ## E = mgy_{max}##

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E = total energy

When the particle is at ##y_{max}## then all the energy = U, thus ## E = mgy_{max}##
Sorry, missed out a crucial word ("not") in post #14.
So, yes, ## E = mgy_{max}##.

What dynamic energy equation does that give you (instead of the one at the end of post #13), and what equation connects ##\dot y, \dot x## ( it is not ##\dot y=2A\dot x##)?

happyparticle
What dynamic energy equation does that give you (instead of the one at the end of post #13), and what equation connects ##\dot y, \dot x## ( it is not ##\dot y=2A\dot x##)?

## \dot{y}^2 + \dot{x}^2 = 2g(y_{max} - y)##

##\dot{y} = 2Ax##

##\dot{x} = \frac{1}{2\sqrt{A}\sqrt{y}}##

sorry, I can't find a connection between ##\dot{x},\dot{y}##

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##\dot{y} = 2Ax##
sorry, I can't find a connection between ##\dot{x},\dot{y}##
That equation is wrong. You have differentiated the LHS wrt t but the RHS wrt x. How do you differentiate Ax2 wrt t?
The relationship I am asking for is the correct form of that.

happyparticle
I'm sorry...
##\dot{y} = A\dot{x}^2##

##\dot{y}^2 + \frac{\dot{y}}{A} = 2g(y_{max} - y)##

I can't have ##\sqrt{2Ag}##

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I'm sorry...
##\dot{y} = A\dot{x}^2##
Still wrong.
Use the chain rule.

Delta2
happyparticle
A is a constant
##\frac{dy}{dt} = A \frac{d(x^2)}{dt}##

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A is a constant
##\frac{dy}{dt} = A \frac{d(x^2)}{dt}##
Yes, but what do you get when you apply the chain rule to the RHS?
You've so far tried ##2Ax## and ##A\dot x^2##.
(You are familiar with the chain rule?)

happyparticle
(You are familiar with the chain rule?)

I think so, but I don't see how to use the chain rule here.
I mean, I can use the chain rule if I have something like ##\sqrt{(x^2y-xy)}##

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I think so, but I don't see how to use the chain rule here.
I mean, I can use the chain rule if I have something like ##\sqrt{(x^2y-xy)}##
##\frac d{dt}f(x(t))=\frac d{dx}f.\frac d{dt}x##.

happyparticle
x is a function?

x = x(t) ?

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x is a function?

x = x(t) ?
Yes. In your problem x and y both depend on t, and there is a fixed relationship between x and y, so you can think of it as y depends on x depends on t.

Delta2
happyparticle
Alright,

##f(x) = Ax^2##

##\frac{d}{dx}f = 2Ax ##
##\frac{d}{dt}x = \dot{x}^2##

##2Ax \cdot \dot{x}^2##

Is that right?

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##\frac{d}{dt}x = \dot{x}^2##
Inventive.

happyparticle
I though if ##f(x) = Ax^2, x(t) = x^2##

why not ##x^2## ?

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I though if ##f(x) = Ax^2, x(t) = x^2##
What if I were to define another function g(x)=x3? Would that suddenly make x(t) equal x3?
No, x(t) just means that x is some function of t. It doesn't tell you what the function is.
##\frac{d}{dt}x=\dot x##.

happyparticle
Alright, so I got

##2Ax \cdot \dot{x} = 2g(y_{max}- y)##

I still don't see how to get
##\omega = \sqrt{2gA}##

##t = \int\frac{x}{\frac{g(y_{max}-y)}{Ax}} dx##

##\omega = \frac{2\pi}{t}##

##\omega = 2\pi \cdot \frac{g}{Ax^2}##

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Alright, so I got

##2Ax \cdot \dot{x} = 2g(y_{max}- y)##
Have you studied dimensional analysis? It's a powerful way to check your equations.
In the present case we can say x and y have length dimension (L). We write that as [x]=[y]=L.
To be consistent, A has dimension 1/L, so that ##y=Ax^2## works out correctly. In the notation: [A]=L-1,
Derivatives are like division: ##[\dot x]=LT^{-1}##, etc.
In your equation above, on the left you have
##L^{-1}L.LT^{-1}=LT^{-1}##, i.e. velocity.
On the right: ##LT^{-2}.L=L^2T^{-2}##, i.e. a velocity squared.
So the equation has to be wrong. This technique would have picked up some of your previous errors.

Anyway, you now have correctly that ##\dot y=2Ax.\dot x##.
Plug that into

##\dot{y}^2 + \dot{x}^2 = 2g(y_{max} - y)##

Hamiltonian and Delta2
happyparticle
Have you studied dimensional analysis? It's a powerful way to check your equations.
No, I don't. Maybe that's why I don't fully understand. However, it should be a way to get the answer by using the small oscillations and the simple harmonic motion.

I was trying to get something in this form $$\dot{\theta} = \sqrt{2Ag}[elliptic integral]$$ or $$\ddot{\theta} + \frac{g}{l} sin \theta = 0$$

Using F = Ma or E = K+U

However, as I said I don't know how to deal with a parabola.

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No, I don't. Maybe that's why I don't fully understand. However, it should be a way to get the answer by using the small oscillations and the simple harmonic motion.

I was trying to get something in this form $$\dot{\theta} = \sqrt{2Ag}[elliptic integral]$$ or $$\ddot{\theta} + \frac{g}{l} sin \theta = 0$$

Using F = Ma or E = K+U

However, as I said I don't know how to deal with a parabola.
Well, you seem to be prone to algebraic errors, so I strongly recommend you study dimensional analysis. It's not hard.
Meanwhile, please proceed as I instructed at the end of post #32. You need to get to an equation in which the only variables are x and its time derivatives.

happyparticle and Delta2
happyparticle