When a spring of constant k is cut into 2 equal parts then new spring constant of both the parts would be?

Question : A uniform spring with a constant 120 N/m is cut into two pieces, one twice as long as the other. What are the spring constants of the two pieces?

I know that when a spring is cut into 2 or more pieces of equal length, we apply either$$k_{original} = \frac{EA}{L_{orginal}}=\frac{EA}{n .L_{orginal}}$$ (where n is number of pieces of equal length ) or $$ E = \frac{1}{2}kx^2 $$

to find out what is the new spring constant of each piece of spring. However in this question, where one spring is cut in $\frac{2L}{3} $and other in $\frac{1L}{3}$ how shall I find out their spring constants? What I did was :

$$E = \frac{1}{2}k(\frac{2L}{3})^2$$ and $$E = \frac{1}{2}k(\frac{L}{3})^2 $$ I am stuck because I realize that energy would not be the same in each of the two pieces so they can not be equated with energy of undivided spring.

Text Solution

Solution : Let us consider that the spring elongates by x when a force F is applied on it. So, the force constant of the spring `K=F/x`. <br> Now, if the spring is cut into two equal parts. Then on the application of the same force F, each part of the spring will elongated by `x/2`. <br> `therefore` The force constant of each part ,<br> `k.=F/(x/2)=(2F)/x=2k`