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

[happyparticle]

Alright, so I got

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

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

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

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

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

 

[haruspex]

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

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

 

[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.

 

[haruspex]

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]

Alright, thanks for your help and your patience.

 

[Delta2]

Yeap, @haruspex is great and he has very great patience too :D.

 

[kuruman]

This problem fooled me at first into thinking that the potential is harmonic because the potential energy is $U=mgy=mgAx^2$. As such, I reasoned, there must be a clearly defined frequency of oscillations. Then it dawned on me that this is not the case here because the motion is two-dimensional and not restricted only in $x$. A 2d parabolic cup $y=Ax^2$ with gravity along $y$ does not give rise to an amplitude-independent oscillation frequency. The appropriate shape that does that is the cycloid as is known from solving the brachistochrone problem. Therefore, in this case one needs to apply a small angle approximation for oscillations near the bottom of the parabolic cup.

We can approximate the bottom of the cup to a circle of radius $R$, and write the vertical height relative to the bottom, $$y=R-\sqrt{R^2-x^2}=\frac{(R)^2-(R-\sqrt{R^2-x^2})^2}{R+\sqrt{R^2-x^2}}=\frac{x^2}{R+\sqrt{R^2-x^2}}=Ax^2.$$All that remains to obtain the answer in post #1 is to find an approximate expression for $y$ when $x<<R$, solve for $R$ in terms of $A$ and substitute in the expression for the frequency of a pendulum of length $R$.

 

[haruspex]

We can approximate the bottom of the cup to a circle of radius R,

Yes, that's ok, but I don't think it is that helpful. A further approximation will be required.
For the actual parabolic shape I got $(1+4Ax^2)\dot x^2=2gA(X^2-x^2)$, where X is the initial x displacement.
One route from there is to differentiate and approximate for small X, x to obtain the classic SHM ODE. Another is to make the small x approximation immediately to get it into the form $\dot x^2+\omega^2x^2= constant$, then apply a small X approximation,

 

[kuruman]

Yes, that's ok, but I don't think it is that helpful. A further approximation will be required.

I don't see where a further approximation is required. The expression $$y=\frac{x^2}{R+\sqrt{R^2-x^2}}$$ is exact for a circular path. The only approximation comes in when one says that for $x<<R$, one gets $y\approx\dfrac{x^2}{2R}$, still for a circular path. Now it is always true that, for the actual parabolic path, $y=Ax^2$ for any $x$. Therefore for small $x$ (in the parabolic path) one can match the parabola to a circle if one sets $$\frac{x^2}{2R}=Ax^2~\Rightarrow~R=\frac{1}{2A}.$$ For a pendulum following a circular path of radius $R$ and in the small angle approximation $\omega=\sqrt{g/R}$. The answer follows.

 

[haruspex]

The only approximation comes in when one says that for x<<R,

That's the one.
I am saying I don't see the benefit in approximating it as circular first. There is no great difficulty in writing the exact ODE for the parabola and then doing the small displacement approximation.

 

[kuruman]

That's the one.
I am saying I don't see the benefit in approximating it as circular first. There is no great difficulty in writing the exact ODE for the parabola and then doing the small displacement approximation.

Difficulty is in the eye of the solver. The benefit I see is that the method I outlined above, although less elegant than yours, makes the explanation more accessible to a beginner. I'll leave it at that because it's not a point worth belaboring.