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Period of mass M hanging vertically from a spring

  • Thread starter yasar1967
  • Start date
73
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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
 

Answers and Replies

18
0
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]
 
73
0
thank you.
 

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