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Compound Harmonic Motion

  1. Feb 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Spring.jpg

    Determine the motion of this mechanical system satisfying the initial conditions :-
    y1(0) = 1
    y2(0) = 2
    y1'(0) = -2*sqrt(6)
    y2'(0) = sqrt(6)

    Hint : there are 4 different methods you can use to solve this problem. They all give the same exact result.

    I need to know what y1(t) and y2(t) are, please


    2. Relevant equations
    I know the shm equations- are they valid here?


    3. The attempt at a solution

    total k = 1/(1/k1+1/k2)

    total m = m1 + m2

    Help!
     
  2. jcsd
  3. Feb 14, 2014 #2
    You could start with a free body diagram for each mass and the corresponding equations.
     
  4. Feb 14, 2014 #3
    Hmm do I need a diagram to solve???
     
  5. Feb 15, 2014 #4
    Strictly speaking no, but it usually helps. What you need is all the forces on each mass.
     
  6. Feb 16, 2014 #5
    Any ideas?
     
  7. Feb 16, 2014 #6
    Some ideas have been given. But we are not seeing any effort on your part.
     
  8. Feb 16, 2014 #7
    I got another help from a friend, here is what he told me...

    The calculations in your way will be rather long so I use the Laplace transformations. Are you familiar with that?
    The first and best is Laplace. The second, long one, is to add different sin AND cos curves. The first approach is to give the y1- and y2-equations right. Your approach is wrong. Then put the initial conditions right. One is wrong for you. Then determine w.
    When you done that one can write the curves in a graph. I have done that in my solution to control my equations solution. As you suggest you can do it any way if you have the initial conditions right (I believe).
    This is a really hard problem for one who is not familiar and the Laplace as a working tool.

    I am not familiar with Laplace. Are you? Here is my idea...

    Upper mass:
    Upper mass, upward spring force -> -k1*y1 = -3*y1
    upper mass, downward spring force -> k2*(y2-y1) = 2*(y2-y1)
    the change in lenght of the upper spring is y1, and in the lower spring is y2 - y1
    Eq1 --> y1" = -5*y1 + 2*y2
    Eq2 --> y2" = 2*y1 - 2*y2

    I think the solutions may be

    1. Reduction to a fourth-order differential equation.
    2. Triangularization using operators.3. Substitution of Y(t) = X*exp(wt).3. Substitution of Y(t) = X*exp(wt).
    4. Diagonalization.
     
    Last edited: Feb 16, 2014
  9. Feb 20, 2014 #8
    Any ideas?
     
  10. Feb 20, 2014 #9
    I am sorry, somehow I missed your previous message. Your method is essentially correct, but you forget there is also gravity.
     
  11. Feb 20, 2014 #10
    I am told gravity pulls on the weights to start the springs moving, but you don't need to deal with gravity in the calculations.
     
  12. Feb 21, 2014 #11
    Do you mean that ##y_1## and ##y_2## are measured from the equilibrium condition?
     
  13. Feb 21, 2014 #12
    ideal springs, point masses cannot collide, y1 and y2 are the distances of the bottom end of the springs from the top, so that the length of the second spring is y2-y1. As for gravitational effects, gravity pulls on the weights to start the springs moving, but you don't need to deal with gravity in your calculations.
     
  14. Feb 21, 2014 #13
    This statement: "gravity pulls on the weights to start the springs moving, but you don't need to deal with gravity in your calculations" does not make a whole lot of sense. Still, if you are told to ignore gravity, then you all you need to do is solve your equations.
     
  15. Feb 21, 2014 #14
    Thats where I need the help :)
     
  16. Feb 21, 2014 #15
    You listed four methods. Surely at least one of them should work.
     
  17. Feb 21, 2014 #16
    That's the problem of where I'm stuck. Help a brother out?
     
  18. Feb 22, 2014 #17
    Since you already know that the system will be undergoing harmonic motion, you can let ##y_1 = a \sin \omega t, y_2 = b \sin \omega t ##, and solve for ##a##, ##b## and ##\omega##. There will be two values for ##\omega##, and for each one, a pair of ##a## and ##b##. So you will have $$ (y_1^{(1)}, y_2^{(1)}) = (a_1 \sin \omega_1 t, b_1 \sin \omega_1 t) $$ and $$ (y_1^{(2)}, y_2^{(2)}) = (a_2 \sin \omega_2 t, b_2 \sin \omega_2 t) $$ The general solution is then $$ A (y_1^{(1)}, y_2^{(1)}) + B (y_1^{(2)}, y_2^{(2)}) $$
     
  19. Feb 23, 2014 #18
    The only question remaining is how the heck do I solve for a, b, w1 and w2?
     
  20. Feb 24, 2014 #19
    You should obtain a system of to equations linear in ##a## and ##b##. Two coefficients of the linear system will depend on ##\omega^2##. The system will have the trivial solution ##a = b = 0##. You need to find a non-trivial solution. For that, you must equate the determinant of the system with zero, that will give you an equation for ##\omega^2##. Solve it, you will end up with two values for ##\omega^2##. Then substitute each one into the system and obtain ##a## and ##b##.
     
  21. Mar 26, 2014 #20
    A(y(1)1,y(1)2)+B(y(2)1,y(2)2) is what we have...so a and b must equal 0? I should be able to plug ANY numbers in to each a and b and be able to have them equal correct? Or is there a correct way to do this...?
     
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