Question

Spring of force constant $k$ is cut into springs of length ratio $1:2:3$. They are then connected in series and the new force constant is $k^{\prime }.$ If they are connected in parallel, then the force constant is $k^{\prime \prime }.$ Find $k^{\prime }:k^{\prime \prime }$

Answer 1

Let $l$ be the complete length of the spring.

Length when cut in ratio, $1:2:3$ are $\frac{l}{6},\frac{l}{3}\text{and}\frac{l}{2}$

Spring constant $\left(k\right)\propto \frac{1}{\text{length}\left(l\right)}$

Spring constant for given segments , $k_1=6k,k_2=3k\text{and}{k}_3=2k$

When they are connected in series , $\frac{1}{k^{\prime }}=\frac{1}{6k}+\frac{1}{3k}+\frac{1}{2k}$

$\frac{1}{k^{\prime }}=\frac{6}{6k}=\frac{1}{k}$

$\therefore$ force constant $k^{\prime }=k$

And when they are connected in parallel , $k^{\prime \prime }=6k+3k+2k=11k$

The ratio's

$\frac{k^{\prime }}{k^{\prime \prime }}=\frac{k}{11k}i.e.,k^{\prime }:k^{\prime \prime }=1:11$

Hence, option (b) is the correct answer.