Series and parallel springs

Equivalent spring

The following table gives formulas for the spring that is equivalent to an ensemble (or system) of two springs, in series or in parallel, whose spring constants are ${\displaystyle k_{1}}$ and ${\displaystyle k_{2}}$.^[1] (The compliance ${\displaystyle c}$ of a spring is the reciprocal ${\displaystyle 1/k}$ of its spring constant.)

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Quantity	In Series		In Parallel
Equivalent spring constant	${\displaystyle {\frac {1}{k_{\mathrm {eq} }}}={\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}}$		${\displaystyle k_{\mathrm {eq} }=k_{1}+k_{2}}$
Equivalent compliance	${\displaystyle c_{\mathrm {eq} }=c_{1}+c_{2}}$		${\displaystyle {\frac {1}{c_{\mathrm {eq} }}}={\frac {1}{c_{1}}}+{\frac {1}{c_{2}}}}$
Deflection (elongation)	${\displaystyle x_{\mathrm {eq} }=x_{1}+x_{2}}$		${\displaystyle x_{\mathrm {eq} }=x_{1}=x_{2}}$
Force	${\displaystyle F_{\mathrm {eq} }=F_{1}=F_{2}}$		${\displaystyle F_{\mathrm {eq} }=F_{1}+F_{2}}$
Stored energy	${\displaystyle E_{\mathrm {eq} }=E_{1}+E_{2}}$		${\displaystyle E_{\mathrm {eq} }=E_{1}+E_{2}}$

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Derivations of spring formulas (equivalent spring constant)

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Equivalent Spring Constant (Series)

When putting two springs in their equilibrium positions in series attached at the end to a block and then displacing it from that equilibrium, each of the springs will experience corresponding displacements x1 and x2 for a total displacement of x1 + x2. We will be looking for an equation for the force on the block that looks like: ${\displaystyle F_{b}=-k_{\mathrm {eq} }(x_{1}+x_{2}).\,}$ The force that each spring experiences will have to be same since otherwise, the springs would buckle. Moreover, this force will be the same as Fb. This means that ${\displaystyle F_{1}=-k_{1}x_{1}=F_{2}=-k_{2}x_{2}=F_{b}.\,}$ Working in terms of the absolute values, we can solve for ${\displaystyle x_{1}\,}$ and ${\displaystyle x_{2}\,}$: ${\displaystyle x_{1}~=~{\frac {F_{1}}{k_{1}}}\,,\qquad x_{2}~=~{\frac {F_{2}}{k_{2}}}}$, and similarly, ${\displaystyle x_{1}~+~x_{2}~=~{\frac {F_{b}}{k_{\mathrm {eq} }}}}$. Substituting ${\displaystyle x_{1}\,}$ and ${\displaystyle x_{2}\,}$ into the latter equation, we find ${\displaystyle {\frac {F_{1}}{k_{1}}}~+~{\frac {F_{2}}{k_{2}}}~=~{\frac {F_{b}}{k_{\mathrm {eq} }}}}$. Now remembering that ${\displaystyle F_{1}~=~F_{2}~=~F_{b}}$, we arrive at ${\displaystyle {\frac {1}{k_{\mathrm {eq} }}}={\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}.\,}$

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Equivalent Spring Constant (Parallel)

Both springs are touching the block in this case, and whatever distance spring 1 is compressed has to be the same amount spring 2 is compressed. The force on the block is then: ${\displaystyle F_{b}\,}$ ${\displaystyle =F_{1}+F_{2}\,}$ ${\displaystyle =-k_{1}x-k_{2}x\,}$ So the force on the block is ${\displaystyle F_{b}=-(k_{1}+k_{2})x.\,}$ Which is how we can define the equivalent spring constant as ${\displaystyle k_{\mathrm {eq} }=k_{1}+k_{2}.\,}$

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Compressed Distance
In the case where two springs are in parallel it is immediate that: ${\displaystyle x_{1}=x_{2}\,}$ ${\displaystyle F_{1}=-k_{1}x_{1}}$ and ${\displaystyle F_{2}=-k_{2}x_{2}.\,}$	In the case where two springs are in series, the force of the springs on each other are equal: ${\displaystyle F_{1}=F_{2}\,}$ ${\displaystyle -k_{1}x_{1}=-k_{2}x_{2}.\,}$ From this we get a relationship between the compressed distances for the in series case: ${\displaystyle {\frac {x_{1}}{x_{2}}}={\frac {k_{2}}{k_{1}}}.\,}$

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Energy Stored

For the series case, the ratio of energy stored in springs is: ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {{\frac {1}{2}}k_{1}x_{1}^{2}}{{\frac {1}{2}}k_{2}x_{2}^{2}}},\,}$ but there is a relationship between x1 and x2 derived earlier, so we can plug that in: ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}\left({\frac {k_{2}}{k_{1}}}\right)^{2}={\frac {k_{2}}{k_{1}}}.\,}$ For the parallel case, ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {{\frac {1}{2}}k_{1}x^{2}}{{\frac {1}{2}}k_{2}x^{2}}}\,}$ because the compressed distance of the springs is the same, this simplifies to ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}.\,}$ GATO Ignace notes about spring associations

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