@tomleslie @acer

I agree with your answers and I think the OP is confused and needs to revisit beam vibration theory. The beam has one rigid body mode (frequency = 0) since the vertical movement is not prevented and the other frequencies are as you found.

1. Shouldn't the acceleration term have c*v not just v? I assume this is a viscous damping term.

2. Why do you expect the condition x = v = 0 to occur? Why not solve the system without events and look at the solution to see if this condition is possible.

@mohammad1369 

I don't know anything about DQM, whatever it stands for. I think your problems are theoretical and not with Maple. I can only suggest that you try solving a simple problem for which you know the answer and go from there.

Good luck.

@mohammad1369 

If you want the frequencies look at a standard book on elasticity or plate theory depending on what mode you want. Exact solutions exist for simple cases. If your case is more complicated find an appoximate solution using Rayleigh-Ritz or finite elements. Textbooks discuss how to do this and you don't need to go through the derivation via Hamilton's principle since this is done by the textbooks and they explain how to set up the approximate equations.

@mohammad1369 

I agree with Tom that this has become a guessing game. I think you should go back to the basic theory of the calculus of variations and start again, following my suggestions. There is no need for what you are doing.

@tomleslie 

The OP mentions the calculus of variations, so I guess he wants to use integration by parts to transform the derivatives of the variational functions to non-derivative terms, although he hasn't told us which of the functions they are supposed to be. Clearer notation like delta_g(x,y) and delta_h(x,y) would help.

If this is what he wants to do, I think it's better to use the calculus of variations abilities of Maple directly or perhaps use the old calcvar package in the Application center.

This is an unusual ODE. May I ask what the physical problem is?

@Rouben Rostamian  

This is excellent. May I ask how you found Lcrit?

@Rouben Rostamian  

This is a fine solution to the problem.

Your conclusion that it's possible for the disk to detach relies on the reaction going to zero. At this stage fy cannot physically become negative and so I presume that the rest of the solution is not valid after this. I wonder if it's possible to provide a solution that handles the constraint fy >= 0?

I guess it's more challenging to set up the equations in this case but it would be interesting to see if Maple could solve it. I would ignore air resistance effects to keep it simple. Unfortunately, I don't know enough about the physics of the problem to attempt this but others may be able to do it.

John

Excellent work John and this is a great illustration of the power of Maple. My only complaint is that your computer is much faster that mine. :)

John

Thanks for the additional information.

I was hoping for this to be done automatically like dsolve does because introducing more variables becomes tedious with many of these types of boundary conditions.

Thanks for this software. It's good to have an alternative to dsolve and the old shooting code from Douglas Meade.

Is it hard to implement mixed boundary conditions?

@asa12 

There are three solutions (using Maple 2017):

AA := {-m*diff(diff(x(t),t),t), m*diff(x(t),t) = K[1], -1/2*m*diff(x(t),t)^2-1/R^4 = K[2]};

The secod solution follows from integrating the basic solution (first solution term = 0). The third solution says that if Action is a constant, the EL equation is zero = zero. See a book on calculus on variations for the basic theory. The value of R is irrelevant for the first two solutions, unless it comes up in your boundary conditions.

Which solution you want is up to you to decide as I don't understand the rest of what you are saying.

The eigenvalue problem should have a minus sign for the mass term in your post. The eigenvectors can be multiplied by an arbitrary constant. Did you check if the Maple values are just the book values times a constant?

From what you have presented, wouldn't it be easier to use vectors?

You can then define p1:=<1.2,1.4>; etc and use standard vector algebra to get what you want. p1 can be thought of as the position vector from the origin to the point you are interested in.

This also works in 3d where you can define p1:=<x,y,z>; etc.

Note: This is just a suggestion and isn't meant to detract from the excellent answers you have been given by other users.