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Hooke's Law of a spring with a mass

  1. Aug 5, 2013 #1

    bgq

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    Hi,

    I know that for a massless spring the tension is T = kx, where k is a constant and x is the compression or elongation.

    What is the expression of the tension if the spring has a mass m?

    Thanks in advance.
     
  2. jcsd
  3. Aug 5, 2013 #2
    The tension (stress) does not depend on the mass of the spring.

    Explanation: Elastic properties reflect the variation of the system potential energy with strain (spring extension). In an atomistic picture, the potential energy ultimately arises from interactions between electrons and nuclei, which do not depend measurably on the nuclear mass.

    Massless springs are assumed for problems in which the mass hung on the spring is much larger than that of the spring. If this is not true, then the dynamics of the spring-mass system will change, but for reasons other than the tension.
     
  4. Aug 5, 2013 #3

    Filip Larsen

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    The expression for the force is the same, whether the spring is considered massless or not. However, for a mass-spring system (i.e. a spring with uniformly distributed mass ms connected to a load mass m), the effective mass in dynamic relations (like oscillation period) of the combined spring-mass system is m + ms/3. That is, a third of the springs mass is included as part of the load mass in dynamic situations.
     
  5. Aug 5, 2013 #4

    bgq

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    Thanks to the replies, but...

    If the tension remains the same whether the spring is massless or not, then Newton's second law applied to the block alone is the same: -kx = mx'' which does not include the mass of the spring which is in contradiction with the concept of effective mass.

    How can we apply Newton's second law on the system (Block alone) to the case when the spring is not massless?

    Actually I know to obtain the effective mass m + ms/3 from energy consideration; my problem is how to obtain it from Newton's second Law applied to the block.
     
  6. Aug 5, 2013 #5

    Filip Larsen

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    You can consider the spring as a continuous spring such that the displacement from the unloaded equilibrium is distributed along the spring length. This means you have to model the acceleration of the spring as a differential along the spring where the load mass is m+ms in the fixed end and just m in the freely moving end (where the load mass m is attached).
     
  7. Aug 5, 2013 #6

    AlephZero

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    You can't get the effective mass m + ms/3 from Newton's second law, because it is only an approximation. In the energy method you assume the displacement of the spring is the same as if it did not have any mass.

    That is a reasonably good approximation if ms is small compared with m.

    You can write a differential equation for the motion of the spring and find its eigenvalues and eigenvectors. That will give you an more "exact" solution.

    The lowest eigenvalue will correspond to the approximate solution - a flexible heavy spring with a mass on the end has more than one mode of vibration. In fact the differential equation of motion you get using continuum mechanics has an infinite number of eigenvalues and vectors.
     
  8. Aug 6, 2013 #7

    bgq

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    How?

    What is the approximation?

    How can apply Newton's 2nd Law to the block (not the spring)?
     
  9. Aug 6, 2013 #8

    AlephZero

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