pendulum problem question

[`hýsøŕ]

Hai all, so in the usual pendulum problem where you have ... well a pendulum swinging from side to side under gravity, you end up with

 

d^2 θ /dt^2 = -g/l sinθ

 

where θ is the angle from the vertical axis, g is acceleration due to gravity, l is the length of the pendulum and t is time.

 

now usually when you solve this, you have to use the small angle approximation on the sinθ so that the eqn turns into

 

d^2 θ /dt^2 = -g/l θ

 

which is rly easy to solve,

but my question is, ... before you use the small angle approximation, can you solve the equation analytically and get some kind of function θ(t)? I mean clearly it won't be a trivial solution or even an easy one, it'll probably be reaally hard to solve, but is it theoretically possible? or is there some way of proving that it cannot be done, and you have to make approximations? I've thought about this for years and gotten nowhere lol

[studiot]

To do what you ask you need to go back in the analysis to the opoint where the equations of motion are

 

mgcos(&) - T = -ml(d&/dt)²

 

-mgsin(&) = ml(d²&/dt²)

 

Your equation is the second one.

 

Multiply this by (d&/dt) and integrate

 

0.5ml(d&/dt)² = mgcos(&) + Constant

 

Apply boundary condition for when & = 0 and solve.

 

Edit sorry for using & instead of theta but I don't have my usual stuff available.

Edited April 11, 2015 by studiot

[`hýsøŕ]

nono my question isn't how to solve the physical problem, i just want to know if the equation itself, d^2 θ /dt^2 = -g/l sinθ is directly solvable. like suppose you ended up with that equation in some completely different physics problem.

 

since i posted this i tried wolfram alpha and it gives me some new function i've not heard of, namely the jacobi amplitude function, so i guess it does kinda exist and i'd just have to learn more about jacobi functions first ... so sorry this seems kinda redundant now lol

[studiot]

I did tell you how to solve the maths, and therefore that it is solvable.

 

Unfortunately I don't have acess to my usual latex editor so I did my best to present it in a standard way.

[ajb]

since i posted this i tried wolfram alpha and it gives me some new function i've not heard of, namely the jacobi amplitude function, so i guess it does kinda exist and i'd just have to learn more about jacobi functions first ... so sorry this seems kinda redundant now lol

That is right, you can solve it in terms of the Jacobi Amplitude function(http://mathworld.wolfram.com/JacobiAmplitude.html). There is a sketch of the solution on Wikipedia http://en.wikipedia.org/wiki/Pendulum_(mathematics)

 

Physically, the system you describe does not show simple harmonic motion; it only does for small angles.

Edited April 13, 2015 by ajb

[`hýsøŕ]

I did tell you how to solve the maths, and therefore that it is solvable.

 

Unfortunately I don't have acess to my usual latex editor so I did my best to present it in a standard way.

 

you told me how to solve the physical problem with a particular boundary condition, but then the equation you gave would just be the one i posted rearranged slightly.. i mean how do you then solve the one you showed me with

 

0.5ml(d&/dt)² = mgcos(&) + Constant

 

? without a boundary condition that is, for generality.

 

 

@ajb yeah i guess it shouldnt perform s.h.m if you imagine a pendulum hehe, i was planning on trying to solve

 

d^2 θ /dt^2 = -g/l sinθ

 

around θ=pi/2 to see what happens when its at 90 degrees and falls but when i expanded it around pi/2, i get like... 1-θ^2 and this is even harder to solve so i was wondering if there was some general way to do it.

but man these jacobi things are complicated, like from my point of view anyway (i've not really done any elliptic integrals before, im up to partial diff. eqns and fourier transforms and that kinda stuff).

[studiot]

 

Your equation is the second one.

 

Multiply this by (d&/dt) and integrate

 

0.5ml(d&/dt)² = mgcos(&) + Constant

 

 

Doesn't this give the general solution?

 

Boundary conditions are applied to eliminate the arbitrary constant for an ODE, or arbitrary function for PDE.

[studiot]

My description was rather short.

 

You can find all the gory detail of the method in chapter IV (miscellaneous theorems and methods) of

 

Applied Differential Equations, by Relton

 

Formally the substitution of the first derivative (Let d&/dt = p) and working in terms of the substituted variable, p has many applications.

The usual reason for using this method is when one or both of the variables are missing from the equation in direct form.

The method reduces the order of the differential equation by one and in your case allows a solution by separation of variables.

 

Your equation is discussed on page 109.