# Period of nonlinear spring-mass system

I wasn't sure where to post this problem, as it's physics related, but rather advanced in its math content (and it's a problem for my applied math course).

## Homework Statement

Considering a spring-mass system (like http://www.cs.toronto.edu/~faisal/teaching/notes/csc418/faisal/img/sm1.gif" [Broken]), given that the nonlinear spring has force = qx^3, where q is the spring stiffness, what is the period of the oscillation when the mass is released from rest at x_0?

## The Attempt at a Solution

The equation of motion of the system is F=ma=-qx^3, so m*((d^2)x/dt^2)+q*x^3=0. Integrating, I get the energy of the system as (1/2)*m*((dx/dt)^2)+(1/4)*q*x^4.
When released from rest at x_0, the system then has E=(1/4)*q*x_0^4. I haven't found examples of this type of problem anywhere, so any help would be greatly appreciated!

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## Answers and Replies

rock.freak667
Homework Helper
$$m\frac{d^2x}{dt^2}-qx^3=0$$

you simply can't integrate both sides

you would have to form the auxiliary eq'n and use the general solution for complex roots.

I've been shown an example of a nonlinear pendulum model, which gives an answer of something involving the EllipticK integral in Maple, but no information on how to derive that answer. For the linear models, the pendulum and spring-mass are directly related, so I don't know if this helps me in any way.

HallsofIvy
Science Advisor
Homework Helper
rock.freak667's point is that you integrated $m d^2x/dt^2$ with respect to t and $-qx^3$ with respect to x! That's what you can't do!

since the independent variable, t, does not appear explicitely, you could use "quadrature". Let v= dx/dt. Then $d^2x/dt^2= dv/dt= (dv/dx)(dx/dt)= v dv/dx$. The equation reduces to mvdv/dx- qx^3= 0. Now you can integrate with respect to x: $(1/2)mv^2- (q/4)x^4$= Constant. (If you look closely at that, you will see "conservation of energy".) Since v= dx/dt, that is the same as
$$dx/dt= \sqrt{\frac{g}{m}}\frac{x^2}{2}$$
That should be easy to integrate. However, I'm not sure that that will answer your question about the period- in general non-linear spring motion is NOT periodic!

x=displacement
a=amplitude
q=spring constant
m=mass
p=period

The energy stored by the spring for a given displacement is given by the integral of the force over the displacement.

$E_{spring}=\int_{0}^{x} q*x^3 dx$

which evaluates to

$E_{spring}=q*x^4/4$

when the mass is at it's maximum displacement and the mass is not moving all the energy in the system is stored in the spring. Thus the total energy in the system is the energy held by the spring when the displacement equals the amplitude.

$E_{system}=q*x^4/4 , x = a$
$E_{system}=q*(a)^4/4$
$E_{system}=q/4*a^4$

applying conservation of energy, any energy lost by the spring must go to the mass as kinetic energy.

The kinetic energy of the mass is given by

$E_{kinetic of mass}=\frac{1}{2}mv^2$

the energy held by the mass can then be expressed as

$E_{kinetic of mass}=E_{system}-E_{spring}$

after substituting this gives

$E_{kinetic of mass}=(q/4*a^4)-(q/4*x^4)$
$E_{kinetic of mass}=\frac{q}{4}*(a^4-x^4)$

substituting this back into the expression for the kinetic energy of the mass

$\frac{q}{4}*(a^4-x^4)=\frac{1}{2}mv^2$

solving for velocity gives

$v_{mass}=\sqrt{\frac{q}{2m}*(a^4-x^4))}$

The mass could of course be traveling in either direction.

recall that

$v=\frac{dx}{dt}$

thus

$\frac{1}{v}=\frac{dt}{dx}$

in other words the reciprocal of velocity is the time passed per unit of distance.

applying this to the expression for the velocity of the mass gives

$\frac{1}{v}=\sqrt{\frac{2m}{q*(a^4-x^4)}}$

or equivalently

$\frac{dt}{dx}=\sqrt{\frac{2m}{q*(a^4-x^4)}}$

integrating both sides dx gives the time taken to travel between the start and end points of that integration. Integrating between $0$ and $a$ provides the time for a quarter period.

$\int_{0}^{a}\frac{dt}{dx}dx=\int_{0}^{a}\sqrt{\frac{2m}{q*(a^4-x^4)}}dx$

(for some reason MathJax isn't rendering the above expression right. I hope it works for everyone else)

$\int_{0}^{a}\sqrt{\frac{2m}{q*(a^4-x^4)}}dx$

$=\sqrt{\frac{2m}{q}}*\int_{0}^{a}\sqrt{\frac{1}{a^4-x^4}}dx$

I'm not sure exactly how to justify this step but nvertheless it should work

$=\sqrt{\frac{2m}{q}}*\frac{1}{a}*\int_{0}^{1}\sqrt{\frac{1}{1-x^4}}dx$

I have no idea how to approach that definite integral analytically but Wolfram can (definite integral~=1.31103)

working from there

$p/4=\sqrt{\frac{2m}{q}}*\frac{1}{a}*1.33103$

$p=\frac{7.41630}{a}*\sqrt{\frac{m}{q}}$

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rock.freak667
Homework Helper
You sort of bumped a 4 year old thread there :rofl:.