Re: Non-linear springs

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

Ed,

The results you have obtained are a special case of the more general
result that *any* purely position-dependent force produces identical
trajectories for particles that have the same initial position, kinetic
energy, and direction of motion.  (If they start at rest, the trajectory
is, thus, dependent *only* upon the initial position.)  Since the kinetic
energies are identical, the speeds along the trajectory vary inversely as
the square root of the mass and, therefore, the times required to complete
any portion of the trajectory vary directly as the square root of the
mass.

These results are very general and, in and of themselves, have no
implications for your question about whether or not there is a closed form
solution in any given situation.

John

On Tue, 23 May 2000, Ed Schweber wrote:

> Ed Schweber (edschweb@ix.netcom.com)
> Physics Teacher at The Solomon Schechter Day School, West Orange, NJ
> To obtain free resources for creative physics teachers visit:
> http://www.physicsweb.com
>
> Hi all:
>
>     I recently set up an interactive physics simulation involving a spring
> with a quadratic restoring force (F = kx^2, with the proper reversal of sign
> as the mass moves through the equilibrium position).
>
>    As I expected, the period was amplitude dependent. But I was initially
> surprised to find that the period was still proportional to the square root
> of the mass as it would be for a linear spring -suggesting the same
> dependence on sqrt(m/k) as for a linear spring.
>
>    However, I also convinced my self that this makes sense. If m = 1 and k=
> 1, the differential equation is:
>
>                             d2x/dt2 = x^2
>
>    If m = 4 (and k still equals 1), the eq is
>
>                              d2x/dt2 = 4x^2
>
>   Define p = (1/2)t
>
>   By applying the chain rule, the equation for the second spring becomes:
>
>                               d2x/dp2 = x^2
>
>    That is the second spring has the same equation in terms of p as the
> first had in terms of t. Therefore the period of the second spring should be
> the same value in terms of p as the period of the first spring was in terms
> of t. But since t= 2p, the period of the second spring should be twice as
> large in terns of t as the period of the first spring was.
>
>    Since the amplitude is the only other factor that can effect the period
> this seems to suggest that the formula for the period is
>
>                        T = N*sqrt(m/k)*f(A)
>
> where N is a constant and f(A) is some function of the amplitude.
>
>   So my questions are:
>
>            -am I correct so far?
>           -is N = 2*pi as for a linear spring?
>           - can f(A) by determined in closed form?
>
> Thanks for any input
>
> Ed Schweber

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm