A classic of springs

Let $Y$ be the Young's modulus of the material of spring.
$\begin{aligned}Y&=\frac{\frac{kx}{A}}{\frac{x}{l}}\\&=\frac{kl}{A}\\\rightarrow{k}_{1}{l}_{1}&={k}_{2}{l}_{2}\\\huge\color{#3D99F6}{\boxed{{k}_{2}=2k}}\end{aligned}$

If the original spring is stretched a distance $x$ from equilibrium, then the potential energy stored is $PE=\frac{1}{2}kx^2$ . Alternatively, think of the original spring as being made up of the two halves of the spring, connected from end to end. Each half of the spring has a spring constant $k_1$ , to be determined. As the spring is stretched a distance $x$ , each half-spring is stretched a distance $\frac{1}{2}x$ . Each half-spring will have an amount of potential energy stored of $PE_{\text{half}}=\frac{1}{2}k_1(\frac{1}{2}x)^2$ . The amount of energy in the two half-springs must equal the amount of energy in the full spring so:

$PE=2PE_{\text{half}}$

$\frac{1}{2}kx^2=2\frac{1}{2}k_1(\frac{1}{2}x)^2$

$\boxed{k_1=2k}$