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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.

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The total energy will be the same, thus: $ \frac{1}{2}k L ^2 = \frac{1}{2}k_1(\frac{2L}{3})^2+\frac{1}{2}k_2(\frac{L}{3})^2$,

Also, the force between the two spring segments will be the same ($F_1=F_2$) from which you get:

$k_1\frac{2L}{3}=k_2\frac{L}{3}$

From these two equations you can easily find $k_1$ and $k_1$

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In a spring kl is always constant When you divide it in a ratio 1:3 then spring constant of the springs will be in a ratio 3:1

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