### Article about the Springs in series and parallel

### 1. Springs in series

If the spring is connected in series, as in the figure on the side, then:

1. The increase in the length of spring = the increase in length 1 + the increase in length 2

Δy = Δy_{1 }+ Δy_{1 }

2. The force experienced by equivalent spring = the force experienced by spring 1 = the force experienced by spring 2

F_{s }= F_{1} = F_{2}

3. The equivalent spring’s constant (k_{s})

1/k_{s }= 1/k_{1 }+ 1/k_{2}

Sample problem 1:

Two identical springs each have a constant of 100 N / m connected in series. If the spring’s arrangement is given a load so that it increases 4 cm in length, then the increase in the length of each spring is …

Solution:

The total increase in the length of the two springs is 4 cm, therefore the increase in the length of each spring is 2 cm.

### 2. Springs in parallel

If the spring is connected in parallel, as in the figure on the side, then:

1. The increase in the length of the equivalent spring = the increase in the length of spring 1 = the increase in the length of spring 2

Δy = Δy_{1} + Δy_{1 }

2. The force experienced by the equivalent spring = the force that is experienced by spring 1 + the force experienced by spring 2

F_{s} = F_{1 }+ F_{2}

3. The equivalent spring’s constant (k_{p})

kp = k_{1} + k_{2}

Sample problem 2:

Two springs each with constant c arranged in parallel. The spring constant of this arrangement becomes …

Solution:

The equivalent spring’s constant (k_{p}) = c + c = 2c