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Period of mass M hanging vertically from a spring

  1. Apr 9, 2008 #1
    What is the period of this motion: a mass M hanging vertically from a spring, of spring constant k, under the influence of gravity.



    I figured:
    Mg-ky=Ma=Mw^2y


    (w being angular freguency)

    I know the solution(T=2pi * seqroot(M/k)) but how can I get there?? how can I eliminate Mg? Mg=k*ʌy equation leads to nowhere
     
  2. jcsd
  3. Apr 9, 2008 #2
    well say we have the system at rest, then [tex]Mg=kx_0[/tex], so say now we define our equilibrium point to be [tex]L+x_0[/tex] where [tex]x_0[/tex] is the initial extension, and [tex]L[/tex] is the natural length. introducing a small displacement, the spring will then oscillate about [tex]L+x_0[/tex] under the restoring force [tex]-kx[/tex]. then we see that [tex]\ddot{x}=-\frac{k}{M}x[/tex]. this then gives us [tex]\omega=\sqrt{\frac{k}{M}}[/tex] and thus [tex]T=2\pi \sqrt{\frac{M}{k}}[/tex]
     
  4. Apr 10, 2008 #3
    thank you.
     
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