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Re: Test problem



Greetings,

On 12/5 Rick Tarara wrote:
"Actually with the 5 spring set I use in lab, we have to load the springs
with .1-.5 kg in order to unbind the coils (tested by seeing if the hanging
mass will 'bounce' just a bit or by squinting to see if any light can be
seen through the coils). In a Hooke's law/SHM lab, we then consider the
stretching of the spring ONLY from this initial 'unbound' position. That is
the initial force and the initial displacement are taken as zero even
though some mass has been added and the spring has stretched. Masses are added, displacements measured, and EXTREMELY linear plots are produced and used to find the spring constants. Then a kilogram or more is hung from the springs, put into SHM, periods measured and compared to T = 2pi(m/k)^.5 with m = m-hanging + ½ m-spring. It is the one lab of the semester where there is really good agreement between experiment and theory--a pleasant change! ;-)"

Why is the "effective mass" of the spring one half of its total mass. I have always used 1/3 of the total mass and also gotten very good results (which is not surprising since the spring mass is generally a very small part of the total oscillating system).

I thought that the 1/3 number comes from some complex DE analysis (that I don't do any more ;-))

Stu Leinoff
Adirondack Com Col