Question

When a helical compression spring is cut into two halves, the stiffness of each of the resulting springs will be

• A. unchanged
• B. one-third
• C. one-half
• D. double

Hint:

Spring stiffness is inversely proportional to the number of active turns ($k \propto 1/n$). Halving the length of a spring halves the number of turns, which doubles its stiffness.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The stiffness (or spring constant, $k$) of a spring is defined as the force required to produce a unit deflection.
The stiffness of a helical compression spring is determined by its material properties, wire diameter, coil diameter, and the number of active turns.
Key Formula or Approach:
The formula for the stiffness ($k$) of a helical spring is:
\[ k = \frac{G \cdot d^4}{8 \cdot D^3 \cdot n} \]
Where:
$G$ is the shear modulus of the spring material.
$d$ is the wire diameter.
$D$ is the mean coil diameter.
$n$ is the number of active turns in the spring.

Step 2: Detailed Explanation:
From the stiffness formula, we can see that if the material, wire diameter, and coil diameter remain constant, the spring stiffness is inversely proportional to the number of active turns:
\[ k \propto \frac{1}{n} \]
When a helical compression spring is cut into two equal halves, each of the resulting springs has exactly half the number of active turns as the original spring:
\[ n_{\text{new}} = \frac{n}{2} \]
We can calculate the stiffness of the new spring ($k_{\text{new}}$) as:
\[ k_{\text{new}} \propto \frac{1}{n_{\text{new}}} \]
\[ k_{\text{new}} \propto \frac{1}{(n/2)} \]
\[ k_{\text{new}} = 2 \cdot k \]
Because there are fewer active coils to absorb the applied force, each half-spring becomes twice as stiff (requires double the force to compress by the same distance) as the original spring.
Therefore, the stiffness of each resulting spring is doubled.

Step 3: Final Answer
The stiffness of each resulting spring is double that of the original spring.