Equation of Motion for 2 Springs with Different Constants & Mass Attached

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Homework Help Overview

The discussion revolves around setting up the equations of motion for a system involving two springs with different spring constants and a mass attached to the second spring. The original poster seeks assistance in formulating the equations that govern the behavior of this system under the influence of gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on the mass and how to relate the spring constants to the motion of the mass. Questions arise about the effective spring constant and the distance the mass moves under gravitational force.

Discussion Status

Some participants have offered insights into the relationships between the forces exerted by the springs and the mass. There is an ongoing exploration of how to derive the effective spring constant and the implications of the system's configuration. Multiple interpretations of the problem are being considered, with no explicit consensus reached yet.

Contextual Notes

Participants note the influence of gravitational force on the mass and the need to understand how the two springs interact. There is mention of uncertainty regarding the effective spring constant and the overall behavior of the system.

brad sue
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Please , I need to set up the equation for two springs.

The first one is attached to a ceiling and has a constant k. The second one is attached at the tail of the first one and has a spring constant k'.

If a mass m is attached to the second spring, How can I set up the equation for the system?
 
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the bob will be pulled downward by Earth's gravity.
It will accelerate downward until the Tension in spring #2 is equal to mg .
How far has the bob moved by now? What is the effective "k"?
 
lightgrav said:
the bob will be pulled downward by Earth's gravity.
It will accelerate downward until the Tension in spring #2 is equal to mg .
How far has the bob moved by now? What is the effective "k"?

well I see what you mean but I don't know the answer.

I think that the bob will move down by mg=-kx----> x=mg/k''
k'' is the new constant that I don't know ...

please help me because I don't understand how the system reatcs...

B
 
Firstly, you should solve the whole system's K .
Briefly, F1 is the first spring's force;
F2 is the second spring's force;
then you get F1 = k*Deta X1
F2 = k'*Deta X2
but if the two springs connected, the force will be the same.
so k*Deta X1 = k'*Deta X2 = F
and Deta X = Deta X1 + Deta X2 = F/k+ F/k' = F(k+k')/kk'
so F=Deta X * kk'/(k+k')
and mX'' = (X-X0) * kk'/(k+k')
and you got it.
good luck :smile:
 

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