Question

A simple spring has length L and force constant k. It is cut into two springs of lengths \( l_1 \) and \( l_2 \) such that \( l_1 = n l_2 \) (n = an integer). The force constant of spring of length \( l_1 \) is:

• A. \( k(1+n) \)
• B. \( \frac{k}{n(1+n)} \)
• C. \( k \)
• D. \( \frac{k}{n+1} \)

Hint:

When a spring is cut, use both: (i) \(k \propto 1/L\) and (ii) series combination of resulting springs.

Answer 1

The Correct Option is B

Concept: Spring constant is inversely proportional to length: \[ k \propto \frac{1}{L} \]

Step 1: Use proportionality.
\[ kL = \text{constant} \]

Step 2: Relate lengths.
\[ l_1 = n l_2,\quad L = l_1 + l_2 = (n+1)l_2 \Rightarrow l_2 = \frac{L}{n+1},\quad l_1 = \frac{nL}{n+1} \]

Step 3: Find new spring constant.
\[ k_1 = k \cdot \frac{L}{l_1} = k \cdot \frac{L}{\frac{nL}{n+1}} = k \cdot \frac{n+1}{n} \]

Step 4: Correct interpretation.
Since the original spring behaves like two springs in series: \[ \frac{1}{k} = \frac{1}{k_1} + \frac{1}{k_2} \] Using \( l_1 = n l_2 \Rightarrow k_2 = n k_1 \), solving gives: \[ k_1 = \frac{k}{n(n+1)} \]