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(Nonlinear Dynamics)Find a Mechanical Analog dx/dt=sin(x).

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data
    What is a mechanical analog for xDOT=sin(x) .?Explain how using this analog, it becomes obvious that x*=0 is an unstable fixed point and that x*=pi is a stable fixed point.


    2. Relevant equations

    dx/dt=sin(x)


    The analytical solution to this problem is
    t=ln[(csc(xinitial)+cot(xinitial)/csc(x)+cot(x)]. But that is not what I want.
    I attempted to think of a circle moving around and measure the distance x from the y-axis of a point on the edge of the circle. Unfortunately this would be an analog for dy/dx=sinx.
     
  2. jcsd
  3. Jun 26, 2011 #2
    Its dy/dt = sin(x). The rotation analog still works; just follow the position of a dot that starts out on the y=0 axis.
     
  4. Jun 26, 2011 #3
    Thanks
    but not quite what I meant.
    If you plot dx/dt on the axis normally labled y and x on the x-axis as usual, if you take x=0.1, sinx will be greater than 0 and therefore dx/dt will be greater than 0 and therefore if a particle starts from x=0.1, it will move more to the right.

    Likewise if a particle starts from x=-0.1, sin(x) will be negative and therefore dx/dt will be negative and therefore the particle will move to the left.

    Therefore the particle will move away from x=0 in both directions.

    But if you take x=pi, a little before x=pi, sin(x) will be positive and therefore dx/dt is greater than 0 and therefore from the left of x=pi the particle will move to the right.

    If you take a point a little to the right of x=pi like x=3.2, then sin(x) will be negative, dx/dt will be negative and the particle will move to the left(from the right.

    The particles movement will be as follows(this follows from what I said before)
    ------<--0-->---->--pi--<----<---2pi--->-->- ect.

    to sum up, from the picture it is clear that for x around 0, the particle will go away from x=0 and hence x=0 is unstable and is therefore an unstable fixed point(it is still a fixed point because sin(x)=0 here and therefore dx/dt=0 here. If a particle was to be exactly at x=0, it would stay there until the slightest disturbance came.

    for x around pi, the particles are moving towards x=pi.

    Therefore without the mechanical analog, it is quite clear that x=0 is an unstable fixed point and x=pi is a stable fixed point.

    My question is 'what is a mechanical analog that is modeled by this'
     
  5. Apr 23, 2013 #4

    Lui

    User Avatar

    Following Strogatz's example in overdamped systems, we can model an overdumped pendulum and try to fit such equation.

    Applying the mechanics laws we can achieve this model that becomes basically (T is for Theta):

    d²T/dt² = -(g/l)*sin(T) - (c/ml²)dT/dt.

    As Strogatz states in his example, "If the viscous damping is strong compared to the inertia term", that is (c/ml²)dT/dt >> d²T/dt², we can say that the dynamics behave like (c/ml²)dT/dt = -(g/l)*sin(T).

    Reducing we have (-mlg/c = a),

    dT/dt = a*sin(T) (Approximately).

    Of course we would never have the exact same equation with this model, since 'a' will never be 1. Another bad item of such model is that it reduce the order of the system and with that neglect the initial velocity constant, so we would have the initial velocity as a function of position... I'm sure that this is not the best example but it's a reasonable approach.
     
    Last edited: Apr 23, 2013
  6. May 12, 2013 #5
    Thanks for your reply. I thought that everyone had forgotten about the question. I think I found a better analog after some time. If [tex]\phi [/tex] is one half the angle subtended by the vertical by an inverted pendulum(see figure attached), and if the inverted pendulum is a uniform rod, with the length of the rod [tex]l = \frac{{3g}}{2}[/tex] then [tex]\ddot \phi = \frac{1}{2}\sin \left( {2\phi } \right)[/tex] since [tex]\sum {\tau = \frac{1}{2}mgl} \sin \left( \theta \right) = \frac{1}{2}mgl\sin (2\phi ) = I\alpha = \frac{1}{3}m{l^2}\ddot \theta = \frac{1}{2}m{l^2}\ddot \phi [/tex] . i.e. [tex]\ddot \phi = \frac{1}{2}\sin \left( {2\phi } \right)[/tex]. If the pendulum starts at [tex]\phi = 0[/tex] , then the first integral of the equation is in fact [tex]\dot \phi = \sin \left( \phi \right)[/tex] . This is easy to verify because by the maximal principle, since[tex]\sin \left( \phi \right) \in {C_\infty }[ - \infty ,\infty ] \subset {C_2}\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right][/tex] , differentiating [tex]\dot \phi = \sin \left( \phi \right)[/tex] once to get [tex]\frac{1}{2}\sin \left( {2\phi } \right)[/tex] does in fact tell us that this is the only first integral of the equations for the pendulum. Thus we have found the mechanical analog.

    The second part of the question is how it becomes obvious that for this system, [tex]\phi = 0[/tex] is an unstable fixed point. It is obvious because if you let an inverted pendulum go at [tex]\phi = 0[/tex](i.e when the pendulum is upright), then unless you balance the pendulum perfectly upright, even the smallest pertubation will make the pendulum fall. Physically there is a torque of [tex]\frac{1}{2}mgl\sin (2\phi )[/tex] on the system when is nonzero for theta =0. This is not a restoring torque because it is in the same direction as the angular displacement. Thus it is obvious that [tex]\phi = 0[/tex] is an unstable fixed point.

    p.s. I am allowed to give a full solution because it was me who posted the problem two years ago.
     

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  7. Feb 1, 2017 #6
    Thank you very much.
    Unfortunately, it has been eons since I dealt with an inverted pendulum and I doubt I would have ever been able to get the mechanical analog.
     
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