What happens to the spring constant in a parallel and series spring system?

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In a system of two springs attached to a mass, the effective spring constant (keq) for springs in parallel is the sum of their individual spring constants (keq = k1 + k2). This relationship can be confirmed by applying the equations of motion, which yield a second-order differential equation consistent for both the combined system and the equivalent single spring system. The discussion emphasizes the mathematical verification of this principle, demonstrating its application in a horizontal spring-mass setup. Understanding this concept is crucial for analyzing spring systems in physics. The clarification provided was appreciated by the original poster, highlighting the importance of accurate calculations in academic settings.
saksham
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Hello,
I am a Freshman from International University Bremen. I would like to request you to help me out with a small question.
I have a system of springs. Two springs with spring constants k1 and k2. I attach them to the opposite ends of a mass m1 and then attach the other ends of the springs to a fixed point. I have now a horizontal setup of a spring, a mass, then another spring attached to each other.
Could anyone tell me with small mathematical explanation what happens to the spring constant of the system?
 
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This is equivalent to a system of springs in parallel. So, keq=k1+k2. You can verify this by writing the equations of motion for the system as you described above and then for the system with only keq and the mass. You should get m*x(doubledot)+keq*x=0. Where keq=k1+k2.
 
Thanks.

Thank you very much schutte, I am really grateful to you for saving three marks for me in my mid term.
Cheers,
Saksham
 

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