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Extension in a heavy spring

  1. Aug 27, 2011 #1
    A spring of mass [itex]M[/itex] is suspended from the ceiling of a room. Find the extension in the spring due to its own weight if it has a spring constant of value [itex] k [/itex].

    I am getting answer as [itex] \frac{Mg}{2k} [/itex], but the answer given in back of the book is [itex]\frac{Mg}{3k}[/itex]. What I did was :

    Let the natural (in un-stretched position) length of the spring be [itex]L[/itex]. Let us consider an element of length [itex]dx[/itex] at a distance [itex]x[/itex] from the bottom of the spring. Then the spring constant of this small spring is [itex]k_x = \frac{L}{dx} k[/itex]. Tension in this spring is [itex]T_x= \frac{x}{L} Mg [/itex] (due to the weight of the spring below it, neglecting the weight of this small spring). Therefore, the extension in this small spring, [itex] dl = \frac{T_x}{k_x} = \frac{Mg}{k L^{2}} x dx [/itex]. Therefore total extension, [itex] l = \frac{Mg}{k L^{2}} \int_{0}^{L} x dx = \frac{Mg}{2k}[/itex].

    So, where I am wrong; or is the answer in the book wrong ?
     
  2. jcsd
  3. Aug 27, 2011 #2

    Doc Al

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    Staff: Mentor

    I'd say that you are correct. What book are you using?
     
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