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Simple horizontal harmonic oscillator with spring that has a mass.

  1. Feb 17, 2013 #1


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    Consider a block of mass M connected to a spring of mass m and stiffness k horizontally on a frictionless table. We elongate the block some distance, and then release it so that it now oscillates.

    According to the theoretical study using energy methods, we see that the mass of the spring affects the motion of the block.

    But if we apply Newton's 2nd law to the system formed of the block alone, the mass of the spring do not appear in the equation!!

    Here it is: F = Ma then -kx = Ma then a=(-k/M)x where the weight and the normal reaction cancel each other.

    As we see the mass of the spring does not appear in the equation.

    My question is: Does the mass of the spring affect the motion of the block? and if yes, what is wrong in the way I have applied Newton's second law above?

    Thanks in advance.
  2. jcsd
  3. Feb 17, 2013 #2


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    The mass of the spring will affect the force applied by the spring, as it requires a force to accelerate this mass. There is nothing wrong with your formula, it is just not sufficient to evaluate the motion of the system.
  4. Feb 17, 2013 #3


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    The equation "F = -kx" assumes that the spring is stretched uniformly along its length at all times. That is consistent with the assumption that it has no inertia.

    It would be more correct to say the force depends on the axial strain at the end of the spring. But if you ignore the inertia of the spring, the strain is uniform along the length, and is measured by x/L where L is the unstretched length of the spring. So the OP's equation F = -kx is correct if you ignore the inertia, and include the effect of the length L as part of the spring constant k.

    If you include the inertia of the spring in the model, the spring is not uniformly stretched along its length. To model this you need to set up a differential equation describing the motion of all points along the spring, and the "mass on the end" then becomes part of the boundary conditions for the solution of that equation.

    If you include the inertia of the spring, the solution of the equations of motion include vibrations in simple harmonic motion at more than one frequency (in fact, at an infinite number of different frequences), and any linear combination os those solutions - including solutions that will not look like "simple harmonic motion" and may not even be periodic, because the different frequencies for harmonic vibrations might not be in simple integer ratios like 2:1 or 3:2.
  5. Feb 19, 2013 #4


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    Thank you very much
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