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Effective Mass of Spring

by ShakyAsh
Tags: effective, mass, spring
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ShakyAsh
#1
Nov17-12, 06:03 AM
P: 6
Okay, so I've been trying to figure out what would be the time period of spring of mass 'M' & spring constant K which is hung vertically from a ceiling.

Dividing the spring into n equal massless springs of spring constant k=nK which are connected to each other by n small objects of mass m=M/n like this-
_____
n | (k)
m
:
:
3 | (k)
m
2 | (k)
m
1 | (k)
m
Now, extension in first spring
Δx1=mg/k
Similarly,
Δx2=2mg/k
...
Δxn=n mg/k
ΔX(Total)=ƩΔx=mg n(n+1)/2k
Putting values of m and k
ΔX=Mg n(n+1)/2Kn^2

Taking the limit n→∞
ΔX=Mg/2K which corresponds to an effective mass of M/2 rather than M/3 as given in Wikipedia & other websites.

I would be grateful if someone would be able to explain what i am doing wrong.
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Philip Wood
#2
Nov17-12, 10:49 AM
PF Gold
P: 938
I think what you are doing wrong is to assume that the effective mass of the spring for static extension (equilibrium) is also the effective mass for the dynamic case (oscillations).

The oscillations of a spring with distributed mass will be in the form of standing (stationary) waves. The case you are probably interested in is the fundamental mode. The mathematics is more complicated than for a lumped mass on a massless spring.


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