A spring of force-constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of

Step 1: Determine the length of the long piece.

Let the original length of the spring be $L$.

$$L_{\text{long}} = 2 L_{\text{short}}$$ $$L_{\text{long}} + L_{\text{short}} = L$$ $$L_{\text{long}} + \frac{1}{2}L_{\text{long}} = L$$ $$\frac{3}{2}L_{\text{long}} = L$$ $$L_{\text{long}} = \frac{2}{3}L$$

Step 2: Apply the relationship between spring constant and length.

The force constant of a spring is inversely proportional to its length.

$$k \propto \frac{1}{L}$$ $$k \cdot L = \text{constant}$$

Step 3: Calculate the force constant of the long piece.

Let $k_{\text{long}}$ be the force constant of the long piece.

$$k_{\text{long}} \cdot L_{\text{long}} = k \cdot L$$ $$k_{\text{long}} \cdot \left(\frac{2}{3}L\right) = k \cdot L$$ $$k_{\text{long}} = \frac{3}{2}k$$

Answer: The force-constant of the long piece is $\dfrac{3}{2}k$.