Series and parallel springs

The following table gives formula for the spring that is equivalent to a 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.)

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	${\displaystyle E_{\mathrm {eq} }=E_{1}+E_{2}}$	${\displaystyle E_{\mathrm {eq} }=E_{1}+E_{2}}$

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, 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 ${\displaystyle {\frac {1}{k_{\mathrm {eq} }}}={\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}.\,}$

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 ${\displaystyle k_{\mathrm {eq} }=k_{1}+k_{2}.\,}$

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: ${\displaystyle {\frac {x_{1}}{x_{2}}}={\frac {k_{2}}{k_{1}}}.\,}$

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 ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}.\,}$