- #36

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter happyparticle
- Start date

- #36

- #37

- 12,853

- 6,057

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##.

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##.

Last edited:

- #38

- 39,571

- 8,837

Yes, that's ok, but I don't think it is that helpful. A further approximation will be required.We can approximate the bottom of the cup to a circle of radius R,

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,

- #39

- 12,853

- 6,057

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

- #40

- 39,571

- 8,837

That's the one.The only approximation comes in when one says that for x<<R,

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.

Last edited:

- #41

- 12,853

- 6,057

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.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.

Share:

- Replies
- 28

- Views
- 519

- Replies
- 7

- Views
- 310

- Replies
- 3

- Views
- 359

- Replies
- 51

- Views
- 774

- Replies
- 6

- Views
- 956

- Last Post

- Replies
- 5

- Views
- 312

- Replies
- 6

- Views
- 151

- Replies
- 5

- Views
- 372

- Replies
- 1

- Views
- 241

- Replies
- 2

- Views
- 336