Question

Find the effective spring constant of the series-spring system shown in Figure 5.1 .6 when both springs have the spring constant $\mathit{k}$. Give a physical interpretation of this result.

Short answer

$k_{eff}=\frac{k}{2}$

Step-by-step solution

Definition:

Spring Constant or force constant is defined as the applied force if the displacement in the spring is unity.

Interpretation:

Two springs have the same spring constant $k$.

$k_1=k_2=k$

Two springs supporting a single mass are in series, that is, the springs are attached end to end.

Then the effective spring constant of the system is.

$\begin{array}{rcl}{k}_{eff}& =& \frac{k_1{k}_2}{k_1+k_2}\\ & =& \frac{\left(k\right)\left(k\right)}{k+k}\\ & =& \frac{k^2}{2k}\\ & =& \frac{k}{2}\end{array}$

Physical interpretation:

Suppose two springs supporting a single mass $m$are in series and two springs have same spring constant. Then the effective spring constant of the series-spring system is equal to half of the spring constant of each spring.