Question

A spring force of constant k is cut into length of ratio 1:2:3. They are connected in series and the new force constant is k’. Then they are connected in parallel and force constant is k”. Then k’: k” is:
A. 1:14
B. 1:6
C. 1:9
D. 1:11

Answer 1

Hint: To calculate net stiffness or force constant of springs, we will have to find the stiffness of individual springs and then apply the suitable formula for finding net stiffness of all the springs. After finding net stiffness for both cases, we will divide the two, to find our answer.

Complete Step-by-Step solution:
Let us assume the total length of spring to be l and spring constant of spring k. As the spring is cut into length of ratio 1:2:3, length of corresponding parts will be
$${\displaystyle \frac{l}{6}},{\displaystyle \frac{l}{3}},$$and ${\displaystyle \frac{l}{2}}$ respectively.
The product of spring constant and its length always remains constant, even after cutting the spring into different parts. The product of spring constant and length of original spring is $kl$. This product will remain the same for the different parts.
Let us assume the spring constant of the springs of length $${\displaystyle \frac{l}{6}},{\displaystyle \frac{l}{3}},{\displaystyle \frac{l}{2}}$$ to be $k_1,k_2,$ and $k_3$ respectively. Therefore,
$k_1{\displaystyle \frac{l}{6}}=k_2{\displaystyle \frac{l}{3}}=k_3{\displaystyle \frac{l}{2}}=kl$
From the above equation we get
$k_1=6k,k_2=3k,k_3=2k$
Net stiffness (spring constant) of n springs of stiffness $k_1,k_2,....k_n$ in parallel is given by
$k_{net}=k_1+k_2+...+k_n$
In this case the net stiffness of spring with stiffness $k_1,k_2,$and $k_3$ is k’’. So,
$k^{″}=6k+3k+2k=11k$
$k^{″}=11k$
Net stiffness (spring constant) of n springs of stiffness $k_1,k_2,....k_n$ in series is given by
${\displaystyle \frac{1}{k_{net}}}={\displaystyle \frac{1}{k_1}}+{\displaystyle \frac{1}{k_2}}+....+{\displaystyle \frac{1}{k_n}}$
In this case the net stiffness of spring with stiffness $k_1,k_2,$and $k_3$ is k’. So,
${\displaystyle \frac{1}{k^{\prime }}}={\displaystyle \frac{1}{6k}}+{\displaystyle \frac{1}{3k}}+{\displaystyle \frac{1}{2k}}={\displaystyle \frac{1}{k}}$
$k^{\prime }=k$
Therefore $k^{\prime }:k^{″}={\displaystyle \frac{k^{\prime }}{k^{″}}}={\displaystyle \frac{k}{11k}}=1:11$
Hence the correct option is D.

Note: The formula for the net stiffness for springs connected in parallel is identical to the formula of net resistance of resistors connected in series. And the formula for the net stiffness for springs connected in series is identical to the formula of net resistance of resistors connected in parallel.