Find solution
Differential Equation Problem?

Challenging differential equation for a physics problem( higher math needed)?

  • I thought out one problem in physics. But I have hard trouble setting up the equation. The problem is Lets say there is a elevator with a base area of A In a shaft at an height of H . The mass of the elevator is M. We assume that the shaft beneath the elevator is air tight. At beginning the air pressure in the shaft is po. As you may have thought the elevator starts a free fall. The gas in the shaft does not heat in the process. So po*Vo=p*V applies. What I want is a equation that shows how much has the elevator traveled in any given time. So what I thought was that. The acceleration at any given time is given by M*a=M*g-p*A where p is po*A*H=p*A*(H-x) p=po*H/(H-x) a=g - po*H*A/(M*(H-x)) So I also thought that a=x'' second derivative of x So I get x''=g - po*H*A/(M*(H-x)) Which must be solved to find a function x(t) 1)First question is if this logic is right 2)If not then how should this be done 3) I would also appreciate someone could solve this ( or if he or she gets another equation then solve that). 4) Finally is it possible to lose the assumption that the gas does not heat. I mean would the equation then be solvable. I hope there is someone out there who can solve this

  • Answer:

    Assume T = constant Yes, you have a correct ODE, however, try the energy equation 0 = g h(t) - c/h(t) + 1/2 (dh(t)/dt)^2 h(t) is the height above the ground and "c" is a constant related to mass of 'vator, volume and density and temperature of gas. It is solvable -- but very messy. http://www.wolframalpha.com/input/?i=0=g+x+-+k/x+%2B+1/2+(dx/dt)%5E2 ...... But, you should realize that for small displacements from equilibrium you are going to have a linear relationship for the restoring force which turns the equation into a simple harmonic. i.e. d^2x/dt^2 = - k/m x x = c sin (√k/m * t) Note: F =mg - c/h h = h.e - x h.e = h.equilibrium F = mg + c/(h.e - x) a = c/(m (h.e - x)) for small displacements a ≈ g - (c/(m h.e) + c/(m h.e) x) k = c/h.e a ≈ g - k/m - k x but at equilibrium g = k/m, ergo a ≈ -k/m x ********** curses, jdll is right, the energy equation should be 0 = g h(t) - c/ln (h(t)) + 1/2 (dh(t)/dt)^2 and is unsolvable using std functions. However, the derivation about short movements around equilibrium is correct (I mixed factors from your ODE with the energy equation resulting in an error in the energy equation -- but I got it correct on the derivation for small displacements). //

ANZI at Yahoo! Answers Visit the source

Was this solution helpful to you?

Other answers

I have answered this question in the maths section at the following address: http://answers.yahoo.com/question/index;_ylt=AtGgBMHB7HStfcZQRF6d_nbty6IX;_ylv=3?qid=20100605111411AAh5TGU&show=7#profile-info-jRoqlJPraa Ok, I have checked whether you can integrate if you assume adiabaticity and the answer is yes. (please refer to my answer at the above address for what I mean here) Your equation is now x" = g-Ap0/M(1-x/H)^-gamma and after multiplication by x' and integration you get: 1/2x'^2 = gx +Ap0H/(M(1-gamma))(1-x/H)^(1-gamma) +C with gamma approximately 1.4 C is determined by x'(x=0) = 0, hence Ap0H/(M(1-gamma))+C =0 so finally: 1/2x'^2 = gx +K((1-x/H)^(1-gamma)-1) where I have changed notation: Ap0H/(M(1-gamma)) --> K Now take the square root and separate variables: dx/Sqrt[gx+K((1-x/H)^-.4 -1) = sqrt(2)dt So the question is to find an antiderivative of 1/sqrt(au+(1-u)^.4-1) and I have asked mathematica which answers: -(1/(3 (-1 + au)^3)) Sqrt[-1 + au + 1/(1 - u)^(2/5)] (8 (1 - u)^(1/5) - 4 (-1 + au) (1 - u)^(3/5) - 3 (-1 + au)^2 (-1 + u)) I have no more time to give to this problem, but I think that with this you can at least produce a curve of x as a function of time. @FGR Finding the energy integral is how I started on the other post. However, after multiplying by dx/dt and integrating, you necessarily get a logarithm for the pressure term (Int dx/dt/(H-x)) Of course changing x-->h = H-x does not solve the problem. So I think your energy equation is wrong. The second term should be something like c Log(h(t)) @ANZI I also apparently oversimplified something this time with the integration problem. The constant subtracted under the square root cannot be taken as 1 (Or if it is taken as 1, then there is a coefficient in front of the ^(2/5)) Let me see if there is a primitive

jean-de-la-lune

I am also getting the same differential equation as you. That equation though is difficult to solve analytically. Maple (software) can't find a closed form solution, and that can be taken as a guide. I see also that schmiso has starred this question. He is very good at solving differential equations. If anyone can solve it, he can. But I don't think it can be solved analytically. A numerical approach might be more realistic. Regarding the heating of the gas. If you maintain the ideal gas relation, then PV/T = constant or P = constant * T / (H-x) If it turns out that T (absolute temperature) varies proportionally to (H-x), then and only then does it become easier to solve. Anything else, and you have either the same problem or a more difficult one. *** ANother approach you might consider is to relax the ideal gas relation. If P / P0 = 1 / (1 - x/H), and you know that x is small compared to H (ie the elevator does not go all the way down to the bottom of the shaft), then you can write as an approximation P / P0 = 1 + x/H + O(x/H)^2 = (H + x) / H This will then make the problem solvable. After solving, you can verify whether the assumption x < < H holds true or not.

Dr D

Related Q & A:

Just Added Q & A:

Find solution

For every problem there is a solution! Proved by Solucija.

  • Got an issue and looking for advice?

  • Ask Solucija to search every corner of the Web for help.

  • Get workable solutions and helpful tips in a moment.

Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.