k1xk2/k1+k2

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k1.k2/k1+k2

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It should be noted that when springs are in series then their equivalent stiffeness decreases and when they r in parallel, stiffness increases.so we have to apply a larger force in parallel spring system and lower force in series system.

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nk
Answered On : Aug 5th, 2011
1/k1+1/k2
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1.All Springs in series will experience the same force F applied at the end. (F=F1=F2)
2.Total deflection of this system will be equal to deflection of individual spring (L=L1+L2)
As F=KL
THEREFORE
F1=K1 x L1 , F2=K2 x L2
OR L1=F/K1, L2=F/K2
SO L=L1+L2 WILL BECOME
F/K=F/K1+F/K2
OR 1/K=1/K1+1/K2
OR K=(K1 x K2)/(K1+K2)
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