Question

If three close-coiled and two open-coiled helical springs, each having the stiffness 'k' are connected in series than the overall stiffness is

• A. 5k
• B. k/5
• C. $k/\sqrt{5}$
• D. 6k/5

Answer 1

The correct option is B k/5

It is a series connection so

$\frac{1}{k_{eq}}=\frac{1}{k}+\frac{1}{k}+\frac{1}{k}+\frac{1}{k}+\frac{1}{k}$

$\frac{1}{k_{eq}}=\frac{5}{k}$

$k_{eq}=\frac{k}{5}$