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Effective masses of spring-mass systems

  1. Mar 14, 2009 #1
    I was asked such a question:

    It is known that the effective mass of the spring in this vertical spring-mass system(figure can be viewed by the following Wiki link) is 1/3 of the mass of the spring, for example, if the mass of spring is m, and the mass of the block is M, then the period of the SHO is [tex]2\pi\sqrt{\frac{M+m/3}{k}}[/tex] where [tex]k[/tex] is the spring constant.

    And I found this question can be solved by the energy consideration, for example, in the Wiki demonstration:
    Notice that there is a key step that the velocity of each position of the spring is directly proportional to its length.(which I don't know how to prove it, this is my first question.)


    However, my second question is, Wiki said that the spring of the horizontal spring-mass system has the effective mass 0!
    But I cannot understand why the above argument for solving vertical spring-mass system is not applicable here?!
    This confuses me a lot.


    Moreover, then how about the effective masses of the such following systems? (Please see the attached plots)

    Attached Files:

  2. jcsd
  3. Mar 15, 2009 #2


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    Hi ismaili! :smile:

    This is simple geometry …
    stretched length is proportional to original length, and so d/dt of stretched length is proportional to original length
    Because the weight of the spring does not affect its extension when it's horizontal :wink:
  4. Mar 15, 2009 #3
    Hi Tim, thanks for your discussion!

    "The velocity of the point of the spring is proportional to the distance of the point to the fixed end", this seems to be true only when, say, if we chop the spring into small segments, each segment stretches for the same length.
    This turns out to be the case of the vertical spring-mass system, but now I guess, for other cases, this may not be true.
    But the mass of the spring do play a role in the motion of the system right? My purpose is to calculate the period of such a system, from the energy consideration (in the case that the velocity of the point of the spring is proportional to the distance of that point to the fixed end):
    [tex]E = \frac{1}{2}Mv^2 + \frac{1}{2}\frac{m}{3}v^2 + \frac{1}{2}kx^2[/tex]
    [tex]0 = Mv\frac{dv}{dt} + \frac{m}{3}v\frac{dv}{dt} + kxv=0[/tex]
    [tex]\left(M+\frac{m}{3}\right)\frac{d^2x}{dt^2} + kx = 0[/tex]
    This is basically the method to calculate the period of the system.
    From this argument, even if the spring is horizontal, there exists some velocity function of the distance to the fixed end, say, [tex]v(x)[/tex], then the only difference of the horizontal system to vertical one would be some change of the kinetic energy term of the spring ([tex]\frac{1}{2}\frac{m}{3}v^2[/tex] in the above case.)
    If this is correct, then the effective mass of the horizontal massive spring should not be zero.

    Did I mistake somewhere? :rolleyes:
  5. Mar 16, 2009 #4


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    Hi ismaili! :smile:
    sorry, I'm not understanding this at all :confused:

    the weight cannot matter when the spring is horizontal;

    the mass is already taken into account in the 1/2 kx2

    increase the mass (eg by adding chewing-gum) without affecting the springiness itself, and k changes proportionately :wink:
  6. Mar 16, 2009 #5
    If we consider the total energy of the spring-mass system, the mass of the spring would contribute a kinetic energy term.
    Assume the velocity of the point [tex]x[/tex], where [tex]x[/tex] is the distance of the point to the fixed end, is proportional to the [tex]x[/tex], i.e. [tex]v(x)=\frac{x}{L}v[/tex], where [tex]v[/tex] is the velocity of the block, and [tex]L[/tex] is the length of the spring.
    Then, the kinetic energy of the spring is,
    [tex]\frac{1}{2}\int_0^L\frac{dx}{L}m\left(\frac{x}{L}v\right)^2 = \frac{1}{2}\frac{m}{3}v^2[/tex].
    Then differentiate the total energy equation, we get the equation of motion and we can find that the period of such a system is [tex]2\pi\sqrt{\frac{M+m/3}{k}}[/tex].

    So, as long as the spring is massive, it would contribute to the kinetic energy term, and this affects the motion.

    However, in this method, we basically assumed the potential energy is still [tex]\frac{k}{2}x^2[/tex] which I kinda suspect...
  7. Mar 16, 2009 #6


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    ismaili 1 wikipedia 0

    Hi ismaili! :smile:
    hmm … you seem to be right, and wikipedia is wrong! :biggrin:

    yes, I can't see anything wrong with your reasoning …

    the mass of a spring is usually ignored, because it's so small compared with the mass on the end

    the main wikipedia article, at http://en.wikipedia.org/wiki/Spring_(device)#Simple_harmonic_motion, has it right …
    … but the article on effective mass, at http://en.wikipedia.org/wiki/Effective_mass_(spring-mass_system)#Horizontal_spring-mass_system, that you referred to, is wrong :frown:
    ismaili 1 wikipedia 0 :biggrin:
  8. Apr 20, 2009 #7
    The effective mass is the same in the horizontal configuration as it is in the vertical configuration. The easiest way to verify this is to examine the solution for the vertical configuration and take the limit as g -> 0. Since g is not in the formula for the effective mass...

    By the way, this discussion helps address the my http://www.mathrec.org/old/2001dec/solutions.html" [Broken]: Why is the effective mass for the resonance frequency M + m/3, but the static elongation of the hanging spring depends on M + m/2?
    Last edited by a moderator: May 4, 2017
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