Question

A spring of force constant $k$ is cut into two equal halves. The force constant of each half is
A. $\dfrac{k}{{\sqrt 2 }}$
B. $K$
C. $2k$
D. $\dfrac{k}{2}$

Answer 1

A spring is a pliable object that can be used to store mechanical energy. Spring steel is the most common material used to make springs. There are several spring styles to choose from. Coil springs are often referred to as "coil springs" in ordinary use. When a typical spring is bent or extended from its resting state, it exerts an opposing force that is roughly equal to the length transition (this approximation breaks down for larger deflections).

Complete step by step answer:
Inside the elastic limit of a substance, Hooke's law states that the strain is proportional to the applied stress. As elastic fabrics are strained, the atoms and molecules bend before stress is applied, and then they revert to their original state when the stress is removed.

Hooke's law is generally expressed mathematically as,
$F = - k \cdot x$
The letter $F$ stands for force. $x$ is the length of the expansion. The proportionality constant $k$ is also known as the spring constant. For a spring
${\text{k}}\propto \dfrac{1}{l}$
Suppose spring is cut into two equal halves i.e, ${l_1}\& {l_2}$. Hence,

\Rightarrow {{\text{k}}_2} \propto \dfrac{1}{{{l_2}}} \\\ {\text{ and}} \\\ {\text{ k }} \propto {\text{ }}\dfrac{1}{{{l_1} + {l_2}}} \\\ \Rightarrow {\dfrac{{{k_1}}}{k} = \dfrac{{{l_1} + {l_2}}}{{{l_1}}}{\text{ or }}{{\text{k}}_1} = {\text{k}}\left( {1 + \dfrac{{{l_2}}}{{{l_1}}}} \right)} \\\ $$ $\Rightarrow \dfrac{{{k_2}}}{k} = \dfrac{{{l_1}}}{{{l_2}}} + 1{\text{ }} \\\ \Rightarrow {\text{ }}{k_2} = k\left( {1 + \dfrac{{{l_1}}}{{{l_2}}}} \right) \\\ As{\text{ }}{l_1} = {l_2} \\\ \therefore {k_2} = 2k \\\ $ **So, When length is halved, the spring constant of each half is doubled.** **Note:** The stiffness of the spring is defined as the spring constant. In other words, when the spring's displacement is one unit, the force exerted to induce the displacement is called the spring constant. As a result, it is obvious that the stiffer the coil, the higher the spring constant.