Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A constant force of magnitude F is being applied to the right. When the two springs are connected in this way, they form a system equivalent to a single spring of spring constant k.

Explanation: Spring constant, also known as stiffness constant is the amount of load (force) required to make a deformation of unit length in the size of the spring from its free .

Mathematically given as:

[tex]F=-kx[/tex] ...............(1)

In series the springs are connected back to back consecutively in a single line as shown in figure.

In such a case:

The equivalent elongation [tex]x_{eq}[/tex] is the sum of the respective elongations of each spring.
The equivalent force [tex]F_{eq}[/tex] is equal in each spring.

Mathematically represented as:

[tex]x_{eq}=x_{1}+x_{2}[/tex] ...........................(2)

∵ From eq. (1) & (2) we have,

[tex]x_{eq}=x_{1}+x_{2} [/tex] ∵F=-kx, where negative sign denotes that the restoration force acts in the direction opposite to the applied force.

[tex]\Rightarrow \frac{F_{eq}}{k_{eq}}=\frac{F_{1}}{k_{1}}+\frac{F_{2}}{k_{2}}[/tex]

∵The equivalent force is equal on each spring.

We get the relation as:

[tex]\Rightarrow\frac{1}{k_{eq}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}[/tex]