A spring of spring constant K = 15 N/m is cut into two parts of ratio of length 3 : 1. Find the spring constant of spring with smaller length (in N/m).
Show Hint
A common mistake is to think that the spring constant is directly proportional to the length. Always remember that for a spring made of a uniform material, cutting it makes it stiffer. A shorter spring has a larger spring constant.
Step 1: Understanding the Question:
A spring is cut into two pieces with a length ratio of 3:1. We need to find the spring constant of the smaller piece, given the original spring constant. Step 2: Key Formula or Approach:
The spring constant (K) of a spring is inversely proportional to its length (L). This relationship can be expressed as:
\[ K \propto \frac{1}{L} \quad \text{or} \quad K \cdot L = \text{constant} \]
Step 3: Detailed Explanation:
Let the original length of the spring be \(L\) and its spring constant be \(K = 15\) N/m.
The spring is cut into two parts with lengths \(L_1\) and \(L_2\) in the ratio 3:1.
So, \(L_1 = \frac{3}{3+1}L = \frac{3}{4}L\) and \(L_2 = \frac{1}{3+1}L = \frac{1}{4}L\).
The smaller part has length \(L_2 = \frac{L}{4}\).
Let the spring constant of the smaller part be \(K'\).
Using the relation \(K \cdot L = \text{constant}\), we can write:
\[ K \cdot L = K' \cdot L_2 \]
Substitute the known values:
\[ 15 \cdot L = K' \cdot \left(\frac{L}{4}\right) \]
The length \(L\) cancels out from both sides:
\[ 15 = \frac{K'}{4} \]
\[ K' = 15 \times 4 = 60 \text{ N/m} \]
Step 4: Final Answer:
The spring constant of the smaller part is 60 N/m.
