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Math Help - Double Spring Mass System

  1. #1
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    Double Spring Mass System

    A mass (M) is connected to two oscillating supports with springs (mass in the middle, then single spring on each side connecting to a moving support). The left support oscillates at X_1=Acos(wt) and the right hand support oscillates at L_0+Bwsin(wt) where w is the angular frequency, and L_0 is the separation between the supports when the system is at rest.

    If X(t) is the position of the mass relative to the origin, find a ODE for X(t) and find the location of the mass when at equillibrium/rest. That is, find the position of the mass when its velocity and acceleration are both 0, assuming that A=0 and B=0. Both springs satisfy Hooke's law with constants of k_1 and k_2


    Thats the problem statement.

    I know that this has to be a second order ODE because it's oscillating. Also, I know the form of a mass on the end of a single spring is mx''(t)+kx(t)=0 but I have no idea when it's placed in between.

    How do I start? Just a little nudge in the right direction please.
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  2. #2
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    This is a forced vibration problem with one degree of freedom.

    x''+(k/m)x=(Acos(ot)+L0+Bsin(ot))/m

    xp=Ccos(ot)+Dsin(ot), xp'=-C*o*sin(ot)+D*o*cos(ot), xp''=-C*o^2*cos(ot)-D*o^2*sin(ot).

    k/m=o1^2 angular frequency.

    Substitution of xp and xp'' into diff equation you can find constants and this can determine xp.

    Using initial conditions - velocity and acceleration are both 0, you can find xh now. xh=A1*cos(o1*t)+B1*sin(o1*t), xh'=...,Substituting t=0 you can find xh

    x=xh+xp

    Kind regards!
    Last edited by derdack; April 11th 2011 at 02:14 PM.
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  3. #3
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    Awesome... but we were just told that he forgot to tell us that there is a friction which is proportional to the velocity, where b is for friction. So that means it would be like mx''+bx'+kx=... I don't know what it should equal.
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  4. #4
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    Quote Originally Posted by ShazamV4 View Post
    Awesome... but we were just told that he forgot to tell us that there is a friction which is proportional to the velocity, where b is for friction. So that means it would be like mx''+bx'+kx=... I don't know what it should equal.
    The RHS will be the same as it was in derdack's post. I would note, though, that the standard way of writing the friction term would be to add a -bx' as the friction is opposing the direction of motion.

    -Dan
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  5. #5
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    Quote Originally Posted by derdack View Post
    This is a forced vibration problem with one degree of freedom.

    x''+(k/m)x=(Acos(ot)+L0+Bsin(ot))/m

    xp=Ccos(ot)+Dsin(ot), xp'=-C*o*sin(ot)+D*o*cos(ot), xp''=-C*o^2*cos(ot)-D*o^2*sin(ot).

    k/m=o1^2 angular frequency.

    Substitution of xp and xp'' into diff equation you can find constants and this can determine xp.

    Using initial conditions - velocity and acceleration are both 0, you can find xh now. xh=A1*cos(o1*t)+B1*sin(o1*t), xh'=...,Substituting t=0 you can find xh

    x=xh+xp

    Kind regards!
    What was the benefit of dividing by m? Can't I just leave m in there, get solve for x_h(t) (the homogeneous portion of the ODE), then solve for x_p(t)... or is there a big shortcut/trick that I'm missing here.

    Also, how do you solve for o, in xp(t)
    Last edited by ShazamV4; April 13th 2011 at 11:47 AM.
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