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Hooke's law for vibrating massive spring

  1. Oct 20, 2015 #1


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    Just for curiosity...
    Is Hooke's law valid for a vibrating massive spring ?
    I have done some calculations using both Newton's 2nd Law and the conservation of energy to a horizontal swinging spring connected to a small block in the absence of any friction. I have found that the tension of the spring depends on both the elongation and the acceleration.
    However, the acceleration is multiplied by the mass of the spring, so if the spring is massless, the tension is reduced to T = kx.

    Here is the outline of my work:

    >> I have written the expression of the mechanical energy of the system (block-spring), and then set the derivative to zero. I concluded at the end that I can neglect the mass of the spring if I assume the mass of the block is M + m/3 where m is the mass of the spring and M is the mass of the block. I checked this out on the internet, and I found this conclusion true.

    >> Next I have applied Newton's 2nd Law:
    For a massless spring: T = Ma
    For a massive spring, the tension is T':
    T' = (M + m/3)a (Neglect the mass of the spring and add its third to the block)
    T' = Ma + ma/3
    T' = T + ma/3
    T' = -kx + ma/3 (Note that the tension depends on the acceleration).

    Is my work correct?

    Thank you.
  2. jcsd
  3. Oct 20, 2015 #2

    Simon Bridge

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    Hooks Law is an idealization - it is a useful approximation for many real springs and spring-like situations. So the short answer is "no" - it is not strictly true for any real-World situation.

    You are saying that you can neglect the mass of the spring if M = M + m/3 ??
    Do you mean that if you have a spring mass m and block mass M system, you can model it by an ideal massless spring and a block of mass M+m/3 and get physically the same result for something? (What is it that needs to be the same? Acceleration? The relationship between the restoring force and the extension?)

    If it is on the internet then it must be true!
    (But JIC: please provide the URL.)

    Can't tell. You need to be more careful.
  4. Oct 22, 2015 #3


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    I don't remember the URL, but here is my work.

    Let M be the mass of the spring, L is its length at any time t, and V is the speed of its free end at the time t. Let x be the distance from the fixed end of the spring to any infinitesimal part of the spring, and v is the speed of this part. Let λ be the linear mass density of the spring.

    Assume that the speed of any infinitesimal part of the spring is proportion to x, then v = xV/L.

    The kinetic energy of an infinitesimal part is: dK = 1/2 dm v2 = 1/2 λdx (xV/L)2 = 1/2 M/L (V/L)2 x2dx
    The kinetic energy of the spring is:
    K = ∫dK (from 0 to L) = (MV2)/(2L3) ∫x2dx (from 0 to L) = (MV2)/(2L3) [x3/3] (0 to L) = 1/6 M V2

    Now the total kinetic energy of the system spring-block: K = 1/6 M V2 + 1/2 m V2 = 1/2(m+M/3)V2 which is totally equivalent to consider the spring massless but adding the third of its mass of the block.

    In what follows x is the elongation (or compression) of the spring.

    The total mechanical energy of the system Spring-Block is: E = K + U = 1/2(m+M/3)V2 + 1/2kx2

    Since the mechanical energy is constant then its derivative is zero, so

    (m + M/3) x'' + kx = 0 then mx'' + Mx''/3 + kx = 0
    -Mx''/3 - kx = mx''
    -Ma/3 - kx = ma
    The right side is the mass multiplied by the acceleration of the block, so the left side should equal to the net external force acting on the block which is the tension of the spring.

    Therefore the magnitude of the tension of the spring is T = - kx - Ma/3.
    Last edited: Oct 22, 2015
  5. Oct 22, 2015 #4

    Simon Bridge

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    The relation for U is only if the spring obeys Hooks Law.
  6. Oct 23, 2015 #5


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    Oh, I missed this point. Thanks a lot. I will try to do it again.
  7. Oct 23, 2015 #6

    Simon Bridge

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    Good luck.
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