Period of nonlinear spring-mass system

In summary, the conversation discusses a physics problem involving a spring-mass system with a nonlinear spring. The equation of motion is derived and conservation of energy is used to find the velocity and period of the oscillation. The final solution is a formula for the period in terms of the amplitude, mass, and spring constant.
  • #1
jinteni
2
0
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!
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
[tex]m\frac{d^2x}{dt^2}-qx^3=0[/tex]

you simply can't integrate both sides

you would have to form the auxiliary eq'n and use the general solution for complex roots.
 
  • #3
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.
 
  • #4
rock.freak667's point is that you integrated [itex]m d^2x/dt^2[/itex] with respect to t and [itex]-qx^3[/itex] 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 [itex]d^2x/dt^2= dv/dt= (dv/dx)(dx/dt)= v dv/dx[/itex]. The equation reduces to mvdv/dx- qx^3= 0. Now you can integrate with respect to x: [itex](1/2)mv^2- (q/4)x^4[/itex]= Constant. (If you look closely at that, you will see "conservation of energy".) Since v= dx/dt, that is the same as
[tex]dx/dt= \sqrt{\frac{g}{m}}\frac{x^2}{2}[/tex]
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!
 
  • #5
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.

[itex]E_{spring}=\int_{0}^{x} q*x^3 dx [/itex]

which evaluates to

[itex]E_{spring}=q*x^4/4[/itex]

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.

[itex]E_{system}=q*x^4/4 , x = a[/itex]
[itex]E_{system}=q*(a)^4/4[/itex]
[itex]E_{system}=q/4*a^4[/itex]

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

[itex]E_{kinetic of mass}=\frac{1}{2}mv^2[/itex]

the energy held by the mass can then be expressed as

[itex]E_{kinetic of mass}=E_{system}-E_{spring}[/itex]

after substituting this gives

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

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

[itex]\frac{q}{4}*(a^4-x^4)=\frac{1}{2}mv^2[/itex]

solving for velocity gives

[itex]v_{mass}=\sqrt{\frac{q}{2m}*(a^4-x^4))}[/itex]

The mass could of course be traveling in either direction.

recall that

[itex]v=\frac{dx}{dt}[/itex]

thus

[itex]\frac{1}{v}=\frac{dt}{dx}[/itex]

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

[itex]\frac{1}{v}=\sqrt{\frac{2m}{q*(a^4-x^4)}}[/itex]

or equivalently

[itex]\frac{dt}{dx}=\sqrt{\frac{2m}{q*(a^4-x^4)}}[/itex]

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

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

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

[itex]\int_{0}^{a}\sqrt{\frac{2m}{q*(a^4-x^4)}}dx[/itex]

[itex]=\sqrt{\frac{2m}{q}}*\int_{0}^{a}\sqrt{\frac{1}{a^4-x^4}}dx[/itex]

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

[itex]=\sqrt{\frac{2m}{q}}*\frac{1}{a}*\int_{0}^{1}\sqrt{\frac{1}{1-x^4}}dx[/itex]

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

working from there

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

[itex]p=\frac{7.41630}{a}*\sqrt{\frac{m}{q}}[/itex]
 
Last edited:
  • #6
You sort of bumped a 4 year old thread there :rofl:.
 

What is a nonlinear spring-mass system?

A nonlinear spring-mass system is a physical system composed of a spring and a mass, where the force exerted by the spring is not directly proportional to the displacement of the mass. This results in a nonlinear relationship between the displacement and the applied force.

What is the period of a nonlinear spring-mass system?

The period of a nonlinear spring-mass system refers to the time it takes for the mass to complete one full oscillation. It is affected by factors such as the stiffness of the spring, the mass of the object, and the initial displacement.

How is the period of a nonlinear spring-mass system calculated?

The period of a nonlinear spring-mass system can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant. However, for highly nonlinear systems, this equation may not be accurate and more complex methods may be used.

How does the period of a nonlinear spring-mass system change with different parameters?

The period of a nonlinear spring-mass system is affected by the stiffness of the spring, the mass of the object, and the initial displacement. Increasing the stiffness of the spring or the mass of the object will result in a longer period, while increasing the initial displacement will result in a shorter period.

What are some real-life examples of nonlinear spring-mass systems?

Some real-life examples of nonlinear spring-mass systems include the suspension system of a car, the motion of a pendulum, and the vibrations of a guitar string. These systems exhibit nonlinear behavior due to the varying stiffness of the springs and the mass of the objects involved.

Similar threads

  • Introductory Physics Homework Help
Replies
24
Views
837
  • Calculus and Beyond Homework Help
Replies
2
Views
446
  • Introductory Physics Homework Help
Replies
28
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
257
  • Introductory Physics Homework Help
Replies
16
Views
319
  • Calculus and Beyond Homework Help
Replies
3
Views
474
Replies
4
Views
906
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top