For the spring system given in the figure, the equivalent stiffness is

Concept:

Springs in series:

\(\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} \)

Springs in parallel:

ke = k1 + k2

Calculation:

Given:

Springs with spring constant k and k are connected in parallel,

Let their equivalent spring constant be k1

k1 = k + k

k1 = 2k

Now k1, k and k are in series connection, let their equivalent spring constant is keq

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{k_e}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{2k}} + \frac{1}{{k}} + \frac{1}{{k}} \)

\(\frac{1}{k_{eq}}=\frac{1+2+2}{2k}\)

\(\frac{1}{k_{eq}}=\frac{5}{2k}\)

\(k_{eq}=\frac{2k}{5}\)

k_eq = 0.4 k