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A Forced vibration on mass between two pull springs

  1. Oct 19, 2016 #1
    Consider a mass m with a prestressed pull spring on either end, each attached to a wall. Let k1 and k2 be the pull spring constants of the springs. A displacement of the mass by a distance x results in the first spring k1 lengthening by a distance x(and pulling in the - direction), while the second spring k2 is abbreviating by a distance x (and pulls in the positive direction).

    The equation of motion then becomes:

    Three questions:

    1. Is it correct that upload_2016-10-19_19-58-18.png

    2. If the answer on question (1) is yes, what is ω if k2>k1

    3. Is is possible to resonate with a force F=A.cos(ωt) en what would be the amplitude of this force.
  2. jcsd
  3. Oct 19, 2016 #2


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    What is the resting equilibrium position for each spring individually?
    The way you have it written is at x=0, but you say something about prestressed in your description so it isn't clear.
  4. Oct 19, 2016 #3
    Lets say the resting equilibrium position for the springs are x1 and x2. The reason why they have to be prestressed is because they are pull springs. They can only have a force in one direction. They both are pulling at any time at the mass. When they are not prestressed then, at the position x=0 (with the springs attached) they would be pressed together.
  5. Oct 20, 2016 #4
    I do not think it even matters. The forces of pre-stretched springs eliminate each other at all times.
  6. Oct 20, 2016 #5


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    It should be
    ##ma = (k_1+k_2)x##
    ##ma = (k_1-k_2)x##

    You can show it using the Euler-Lagrange equation.
    ##L = \frac{1}{2}m v^2 - \frac{1}{2}k_2 (x_2-x)^2 - \frac{1}{2} k_1 (x-x_1)^2##
    ##\frac{\partial L}{\partial x} = k_2 (x_2-x) - k_1 (x-x_1) = x (k_1 + k_2) + k_1 x_1 + k_2 x_2##
    If x=0 is the equilibrium position, then ##k_1 x_1 + k_2 x_2 = 0##
  7. Oct 21, 2016 #6
    Why is in this the enegery of both springs negative? Spring 1 is pulling upwards, spring 2 is pulling downwards.
  8. Oct 23, 2016 #7
    Ok, I have discovered the error. The restoring force F2 should also be upwards. Spring 2 also wants to bring the mass to its equilibrium position even if it is a pull spring.
    The question around the negative energy of the spring is also solved.

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