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Factors affecting the spring constant

  1. Apr 15, 2007 #1
    Could anyone please tell me any factors affectin the spring constant k of a spring ?
  2. jcsd
  3. Apr 17, 2007 #2
    I am not entirely sure if this is right, but I think I read somewhere that if you have a spring with spring constant [itex]k[/itex] and you cut it in half, the spring constant then doubles to [itex]2k.[/itex]

    If that is true, can anyone please explain why this is the case?
  4. Apr 17, 2007 #3


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    Welcome to the Forums both of you,

    Corrupt Cannon; The spring constant is the constant of proportionality which appear in Hooke's Law; F = -kx. An alternative expression for Hooke's law is the strain is directly proportional to the stress placed on a object. From this we arrive at an approximation for the spring constant of a solid thus;

    [tex] k = \frac{m}{a^2}\sqrt{\frac{K}{\rho}}[/tex]

    Where [itex]m, a, K \& \rho[/itex] are the mass of a single atom, atomic spacing, bulk modulus and density respectively. However, it should be noted that values of k are determined empirically, i.e. by measurement and forces and displacements, rather than calculation.

    CALCULATOR; Your statement is correct. Suppose we apply a force to a spring with 20 coils, this force results in an extension of 2cm. Therefore, each coil will have extended 0.2cm. Now, let us cut the spring in half, so we now have a new spring with 10 coils. Now suppose we exert a force such that an extension of 2cm is again produced, this means that each coil extends by 0.4cm. Now, according to Hooke's law F = -kx; the force required to extend the spring with 10 coils by 0.4cm is twice as large as the force required to extend the 20 coil spring by 0.2cm (since the extension in the smaller coils is twice that of the larger coil). It therefore follows that the spring constant of the smaller coil must be twice that of the larger coil (since force is proportional to extension).
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