Animated Solution for Physics - Oscillations: If a spring of stiffness $k$ is cut into two parts $A$ and $B$ of length $l_A:l_B=2:3$, then the stiffness of spring $A$ is given by

Imagine you have a long, stretchy rubber band. If you pull it from both ends, it stretches quite easily. Now, imagine holding that same rubber band right in the middle and pulling. It feels much tighter, doesn't it? This simple intuition is the secret to understanding spring constants!

In physics, a spring's stiffness is measured by its spring constant, denoted by $k$. The fundamental rule you must remember is that the spring constant is inversely proportional to the natural length of the spring ($l$). Mathematically, we write this as:

This means if you cut a spring in half, each half becomes twice as stiff as the original spring.

In our problem, we have a spring of stiffness $k$ and length $l$. We are slicing it into two unequal parts, $A$ and $B$, such that the ratio of their lengths is $l_A:l_B=2:3$.

To find the exact length of part $A$, we look at the total number of "parts" the spring is divided into. The ratio $2:3$ means there are $2+3=5$ equal segments in total.

Part $A$ takes up 2 of these 5 segments. Therefore, the length of part $A$ is:

Now we bring back our golden rule: $k\propto \frac{1}{l}$. Because of this inverse relationship, we can set up a ratio comparing the new spring $A$ to the original spring:

Substitute the length of part $A$ that we just found:

The $l$ in the numerator and denominator beautifully cancel each other out. The fraction in the denominator flips, giving us:

And there we have it! The shorter spring $A$ is $2.5$ times stiffer than the original spring. If you were to calculate the stiffness of part $B$, you would find $k_B=\frac{5}{3}k$. Always remember: the shorter the spring, the harder it is to stretch!