Question

A spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the longer piece will have a force constant of

• A. $\frac{2}{3}k$
• B. $\frac{3}{2}k$
• C. $3k$
• D. $6k$

Answer 1

Answer: (B)

$\frac{3}{2}k$

Answer 2

Since, $k\propto\frac{1}{l}$ $\therefore\frac{k_{1}}{k_{2}} = \frac{l_{2}}{l_{1}}$ $\because l_{1} = 2l_{2}$ $\therefore\frac{k_{1}}{k_{2}} = \frac{1}{2}$ But series combination of both cutting portions gives original spring. $\therefore\frac{1}{k} = \frac{1}{k_{1}} +\frac{1}{k_{2}}$ or $\frac{1}{k} = \frac{2}{k_{2}} + \frac{1}{k_{2}} = \frac{3}{k_{2}}$ $\because k_{2} = 3k$ $\therefore k_{1} = \frac{k_{2}}{2} = \frac{3}{2}k$