Question

A spring whose stretched length is $l$ has a force constant $k$. The spring is cut into two pieces of unstretched lengths ${l_1}$ and ${l_2}$ where ${l_1} = n{l_2}$ and $n$ is an integer. The ratio $\dfrac{{{k_1}}}{{{k_2}}}$ of the corresponding force constants, ${k_1}$ and ${k_2}$ will be:
A. $\dfrac{1}{{{n^2}}}$
B. ${n^2}$
C. $\dfrac{1}{n}$
D. $n$

Answer 1

Hint: The spring is divided into two components. The lengths of these components are ${l_1}$ and ${l_2}$. The force constant $k$ is proportional to the length $l$ of the wire. Now, ${k_1}$ and ${k_2}$ are the force constant of lengths ${l_1}$ and ${l_2}$ respectively.

Complete step by step answer:
As given in the question, a spring of length $l$ is divided into two pieces of lengths ${l_1}$ and ${l_2}$ , therefore, we can say,
${l_1} + {l_2} = l$
Now, $k$ is the force constant and is given by
$k \propto \,\dfrac{1}{l}$
$ \Rightarrow \,k = \dfrac{C}{l}$
Here, $C$ is the constant of proportionality and $l$ is the length of the wire.
Now, as the spring is divided into two parts of lengths ${l_1}$ and ${l_2}$.

Therefore, the force constant of the spring having length ${l_1}$ is given by
${k_1} = \dfrac{C}{{{l_1}}}$
Also. The force constant of the spring having length ${l_2}$ is given by
${k_2} = \dfrac{C}{{{l_2}}}$
Now, dividing ${k_1}$ by $k$ , we get
$\dfrac{{{k_1}}}{k} = \dfrac{{\dfrac{c}{{{l_1}}}}}{{\dfrac{c}{{{l_2}}}}}$
$ \Rightarrow \,\dfrac{{{k_1}}}{{{k_{}}}} = \dfrac{c}{{{l_1}}} \times \dfrac{l}{c}$
$ \Rightarrow \,\dfrac{{{k_1}}}{{{k_{}}}} = \dfrac{l}{{{l_1}}}$
$ \Rightarrow \,\dfrac{{{k_1}}}{k} = \dfrac{{{l_1} + {l_2}}}{{{l_1}}}$
$ \Rightarrow \,\dfrac{{{k_1}}}{{{k_2}}} = 1 + \dfrac{{{l_2}}}{{{l_1}}}$
Now, it is given in the question that
 ${l_1} = n{l_2}$
$ \Rightarrow \,\dfrac{{{l_1}}}{{{l_2}}} = n$
Putting the value of $\dfrac{{{l_1}}}{{{l_2}}}$ in the above equation, we get
$ \Rightarrow \,\dfrac{{{k_1}}}{{{k_2}}} = 1 + \dfrac{1}{n} = \dfrac{{n + 1}}{n}$

Now, dividing ${k_2}$ by ${k_{}}$, we get
$\dfrac{{{k_2}}}{k} = \dfrac{{\dfrac{c}{{{l_2}}}}}{{\dfrac{c}{l}}}$
$ \Rightarrow \,\dfrac{{{k_2}}}{k} = \dfrac{c}{{{l_2}}} \times \dfrac{l}{c}$
$ \Rightarrow \,\dfrac{{{k_2}}}{k} = \dfrac{l}{{{l_2}}}$
$ \Rightarrow \,\dfrac{{{k_2}}}{k} = \dfrac{{{l_1} + {l_2}}}{{{l_2}}}$
$ \Rightarrow \,\dfrac{{{k_2}}}{k} = \dfrac{{{l_1}}}{{{l_2}}} + 1$
Putting, the value of $\dfrac{{{l_1}}}{{{l_2}}}$ in the above equation, we get
$ \Rightarrow \,\dfrac{{{k_2}}}{k} = n + 1$
Now, the ratio $\dfrac{{{k_1}}}{{{k_2}}}$ can be calculated by dividing $\dfrac{{{k_1}}}{k}$ and $\dfrac{{{k_2}}}{k}$ as shown below
$\dfrac{{{k_1}}}{{{k_2}}} = \dfrac{{\dfrac{{{k_1}}}{k}}}{{\dfrac{{{k_2}}}{k}}}$
$ \Rightarrow \,\dfrac{{{k_1}}}{{{k_2}}} = \dfrac{{{k_1}}}{k} \times \dfrac{k}{{{k_2}}}$
$ \Rightarrow \,\dfrac{{{k_1}}}{{{k_2}}} = \dfrac{{n + 1}}{n} \times \dfrac{1}{{n + 1}}$
$ \Rightarrow \,\dfrac{{{k_1}}}{{{k_2}}} = \dfrac{1}{n}$
Therefore, the ratio $\dfrac{{{k_1}}}{{{k_2}}}$ is $\dfrac{1}{n}$ .

So, the correct answer is “Option C”.

Note:
Here, we can also divide $K_1$ by $K_2$ directly instead of dividing $K_1$ by $K$ and $K_2$ by $K$ respectively. Also, the length of an ideal spring is equilibrium. Now, if the spring is stretched, the force constant of the spring will be proportional to the increase in the length of the spring and this increase will be an increase in equilibrium length. Therefore, this force will pull each end towards the other.