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Homework Help: Normal Modes

  1. Feb 9, 2010 #1

    bon

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    1. The problem statement, all variables and given/known data

    Ok basically we have two identical coupled pendula - each of length l with bobs of mass m connected by spring with spring constant k.

    so I've shown the eqns of motion are mx'' = -mgx/l + k(y-x) and my'' = -mgy/l - k(y-x)

    The question says: By looking for solutions where x and y vary harmonically at the same angular freqency w, convert these differential equations into two ordinary simultaneous equations for the amplitudes of oscillation X and Y.

    Then it says why do you not expect these eqns to determine the absolute values of X and Y

    2. Relevant equations



    3. The attempt at a solution

    So I think I know the answer to the second q - we lack initial conditions, which are necessary to find absolute values of X and Y..

    Im just not sure how to convert the two DEs into ordinary simulateous eqns for X and Y

    Thanks!
     
  2. jcsd
  3. Feb 9, 2010 #2

    ehild

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    Add the two equations together: you have x and y in the combination x+y. Let it be X. X=x+y

    Subtract one equation from the other: you have the combination y-x. Denote it Y. Y=y-x.


    ehild
     
  4. Feb 9, 2010 #3

    bon

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    Thanks..

    Now have solved this.

    Now trying to solve the harder problem where the masses are not the same - say, they are m1 and m2

    I dont know how to solve this..

    I need to show both decoupling and matrix methods ---

    so I've found the matrix, and know its determinant must = 0, but the expression is just messy...not sure how to proceed...

    How about decoupling methods..what linear combinations of the two equations of motion should i take?

    thanks
     
  5. Feb 9, 2010 #4

    ehild

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    In case of two different masses, one of the new variables can be (y-x) again, as it occurs in both equations. The position of he centre of mass can be the other one:

    [tex]X=\frac{m1*x+m2*y}{m1+m2}[/tex]

    [tex]Y=y-x[/tex].

    The original equation are

    m1x'' = -m1gx/l + k(y-x) and m2y'' = -m2gy/l - k(y-x)

    Add the two equations again to get a new one describing the motion of the centre of mass of the system as a whole, a simple pendulum with mass M=m1+m2.

    Divide the equations by the masses and subtract them from each other, to get equation for y-x. You have the masses in the combination (1/m1+1/m2): this is the reciprocal of the "reduced mass" mu:

    [tex]\mu = \frac{m1*m2}{m1+m2}[/tex]

    and the equation describes SHM of a particle with this reduced mass.



    ehild
     
  6. Feb 10, 2010 #5

    bon

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    Great thanks

    so i get the 2 normal modes to be root g/l and root (g/l + k(m1 + m2) /m1m2 )

    I guess these are right?

    a quick q..

    I'm now meant to find the relative amplitudes..so i know i can write

    x-y = Acos(w1 t + phi 1)
    m1x+m2y/(m1+m2) = Bcos(w2t + phi2)

    where A,B phi1 and phi2 are constants to be determined by initial conditions..
    but how do i find the relative amplitudes?
     
  7. Feb 10, 2010 #6

    ehild

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    Looks good...

    I do not quite understand the question, as the amplitudes can be determined from the original positions and velocities. The question might mean relative amplitudes in a normal mode.

    There are two normal modes, X and Y. If X is "excited" and Y=0, that means that the two masses move together, x=y, so the relative amplitudes are 1/1.
    The other normal mode, Y, is a vibration around the centre of mass which is stationary: X=0.
    That means that m1x+m2y=0--->x/y=-m2/m1.


    ehild
     
    Last edited: Feb 10, 2010
  8. Feb 10, 2010 #7

    bon

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    How though?
    Is there a neat way?

    I can't really see how to do it..

    Thanks again for all your help..
     
  9. Feb 10, 2010 #8

    ehild

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    See my previous post.

    ehild
     
  10. Feb 10, 2010 #9

    bon

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    Thanks so much for your help :)

    As you are good at this, sorry, but Im going to bug you with another Q

    This time I have two vertical springs...

    so top one is attached to a ceiling, then a mass m, which is attached to a spring, which is attached to another mass m (i.e. all in a vertical line)..

    both springs are the same - spring constant k..both masses are mass m

    I'm meant to find the normal mode freqs..

    so im trying to do it by matrix method..

    have two equations mx'' = mg - kx and my'' = mg - k(y-x)

    How do i solve these.? I can't see how to write in the form Ax = 0 where A is some matrix, since mg is neither multiplied my x nor y - if you see what i mean!

    thanks
     
  11. Feb 10, 2010 #10

    bon

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    woops my eq for x is wrong..

    should be mx'' = mg + k(y-x) -kx
     
  12. Feb 10, 2010 #11

    bon

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    sorry - solved this now... you just solve the CF

    to find the ratio of amplitudes do you need to solve PI as well>?
     
  13. Feb 10, 2010 #12

    ehild

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    Sorry, I do not know what is CF and PI.

    Normal mode means that all particles move with the same angular frequency and they have the same phase constant.
    If xo, yo are the stationary values for x and y, the possible vibrational normal mode solutions are: x-xo=Asin(wt+phi) and y-yo=Bsin(wt+phi). Substituting back this functions for x and y in the differential equations, you get an eigenvalue problem: a system of linear equations, which you can solve by matrix method. When the possible values of w have been obtained, plugging one back into the equation, you get the relation between the amplitudes A and B in the normal mode of angular frequency w.

    ehild.
     
  14. Feb 10, 2010 #13

    bon

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    Ah okay this works thanks.
    So then the q asks..why does acceleration due to gravity (g) not appear in these answers?

    Not sure :S?!
     
  15. Feb 10, 2010 #14

    ehild

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    The masses vibrate around their equilibrium positions. In this positions, the springs are strained just as much as to cancel gravity. If the masses move out from their equilibrium by dx and dy, , the force exerted by the springs is -kdl - proportional to the change of length with respect to the length in equilibrium.


    If you use the variables x and y in the meaning you did, then x=xo+A sin(wt +phi), y =yo +B sin(wt+phi). when the bodies are in equilibrium, A=B=0 and the second derivatives are zero. You get the stationary coordinates from here.
    If you replace back the vibrational solutions into the differential equations, xo and yo will cancel with mg and you get a homogeneous system of equations.

    ehild
     
  16. Feb 11, 2010 #15

    bon

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    Thanks - this seems to make perfect sense.

    I'm now just slightly confused about the way I've chose x and y. I chose them to be displacements from the equilibrium position of the mass on the string..

    But then my equation had inhomogeneous terms of g in both the x'' = and y'' = equations.

    So I solved the associated homogeneous equations to find normal frequencies.

    How should i have written it so that the g's would cancel? As displacements from the equilibrium positions of the springs? Bit confused

    Thanks ehild....you are great!
     
  17. Feb 11, 2010 #16

    ehild

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    Both springs are expanded when the masses are in equilibrium, and gravity is balanced by the force of the stretched springs. If you mean x and y as displacement from the equilibrium position, you should not include the term mg.

    If you decide to denote x and y as the distances from the ceiling or from the ends of the unstretched springs, the forces on both masses include gravity. The differential equations are inhomogeneous in this case. Yo certainly know that you solve such equations by solving the homogeneous part first, and adding a particular solution of the inhomogeneous equation to the general solution of the homogeneous one.
    The normal - mode solution of the homogeneous equation is x= A sin(wt+phi), y=B sin(wt+phi). You get a particular solution of the inhomogeneous one by replacing the second derivatives by 0. You get the solution xo, yo.

    If using these xo, yo values, you plug in x=xo+Asin(wt+phi), y = yo+Bsin(wt +phi) into the original inhomogeneous equation, you will see that the xo, yo terms cancel with the mg terms.

    ehild
     
  18. Feb 11, 2010 #17

    bon

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    Thanks SO much for all your help. Understand now.

    Feel embarrassed to ask more, but here it goes..

    two identical coupled pendula - each of length l, bobs of mass m are free to oscillate in same plane..joined by spring, spring const. = k

    i worked out the normal modes to be root g/l and root (g/l + 2k/m)

    then it says at t=0 both pendula are at rest with x=A and y=0. They are then released. Describe subsequent motion..

    What happens? Is just one of the normal modes excited? Which one? How can you tell!

    Thanks ehild! Legendary
     
  19. Feb 11, 2010 #18

    ehild

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    Remember the "normal coordinates" X=(x+y)/2 and Y=y-x.
    Both of them do SHM with their own frequency.

    As X(0)=(A+0)/2 the centre of mass is not in equilibrium at t=0. As Y(0)=(A-0)/2, the distance between the bobs is not equal to the equilibrium value: Both modes are excited.

    ehild
     
  20. Feb 11, 2010 #19

    bon

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    Thanks...ahh approximately how many oscillations does the second pendulum have before the first is oscillating again with its initial amlitude

    doesnt this mean - how many does it have in time t=2pi/ deltaw? how do you work it out?
     
  21. Feb 11, 2010 #20

    bon

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    Is it 66?
     
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