# What is the spring constant in parallel connection and series connection?

Dec 15, 2016

Parallel.

When two massless springs following Hooke's Law, are connected via a thin, vertical rod as shown in the figure below, these are said to be connected in parallel. Spring 1 and 2 have spring constants ${k}_{1}$ and ${k}_{2}$ respectively. A constant force $\vec{F}$ is exerted on the rod so that remains perpendicular to the direction of the force. So that the springs are extended by the same amount. Alternatively, the direction of force could be reversed so that the springs are compressed.

This system of two parallel springs is equivalent to a single Hookean spring, of spring constant $k$. The value of $k$ can be found from the formula that applies to capacitors connected in parallel in an electrical circuit.

$k = {k}_{1} + {k}_{2}$

Series.

When same springs are connected as shown in the figure below, these are said to be connected in series. A constant force $\vec{F}$ is applied on spring 2. So that the springs are extended and the total extension of the combination is the sum of elongation of each spring. Alternatively, the direction of force could be reversed so that the springs are compressed.

This system of two springs in series is equivalent to a single spring, of spring constant $k$. The value of $k$ can be found from the formula that applies to capacitors connected in series in an electrical circuit.

For spring 1, from Hooke's Law

$F = {k}_{1} {x}_{1}$

where ${x}_{1}$ is the deformation of spring.

Similarly if ${x}_{2}$ is the deformation of spring 2 we have

$F = {k}_{2} {x}_{2}$

Total deformation of the system

${x}_{1} + {x}_{2} = \frac{F}{k} _ 1 + \frac{F}{k} _ 2$
$\implies {x}_{1} + {x}_{2} = F \left(\frac{1}{k} _ 1 + \frac{1}{k} _ 2\right)$

Rewriting and comparing with Hooke's law we get

$k = {\left(\frac{1}{k} _ 1 + \frac{1}{k} _ 2\right)}^{-} 1$