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A system of two springs in series and parallel.

Prove that when two springs are attached one at the end of the other, the coefficient of the final spring becomes
1 / (1/k1 + 1/k2 ) where k1 and k2 are the coefficient of the two individual springs.

Then consider two systems of springs, one in which a mass m is attached two the end of two springs which are placed one at the end of the other, another in which m is attached to the two springs in parallel (each spring has coefficient respectively k1 and k2)
Find the frequency of oscillations of mass m in the two cases

Solution This solution is FREE courtesy of BrainMass!

Here is the strategy to solve this problem:

The key point is that when the springs are connected in series and stretched (or compressed) the both exert the same force.

Write down the force each spring exerts (take their displacement from equilibrium lengths as x1 and x2)and equate them.

Express x1 as a function of x2,k1 and k2

The net displacement of a mass connected to the end of the springs is x=x1+x2

The force on the mass must be the same as the force you found earlier, but it can be also written as an *effective constant* times the total displacement

Write down the equation of the forces, substituting now the expression you derived for x1.

Isolate the effective constant.

As for the springs in parallel, now their displacements is the same, and the force exerted on teh mass is the sum of the these forces.

As for teh frequency, once you find the effective spring constant, just substitute it in the motion equation of a simple harmonic oscillator:


And the frequency is w=sqrt(K/M)

If you get stuck, you can always look at the solution.

remark: I assumed you already know how to solve the equation of motion of a simple oscillator.
For the equations of motion solution for a simple harmonic oscillator you might want to check out