Question

A block is attached to a spring and it oscillates with natural frequency \(f_1\). If the spring is cut into two half and only one of the half spring is connected to the block then the frequency becomes \(f_2\). Find \( \frac{f_2}{f_1} \).

• A. \( \sqrt{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \(2\)
• D. \( \frac{1}{2} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The frequency of a spring-block system depends on the spring constant $k$ and mass $m$.
When a spring is cut, the spring constant of the parts changes inversely with length.
Step 2: Key Formula or Approach:
Natural frequency is given by $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$.
The relation between spring constant and length is $k \propto \frac{1}{L} \implies kL = \text{constant}$.
Step 3: Detailed Explanation:
Initially, the frequency of the spring-block system is $f_1 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$.
When the spring is cut into two equal halves, the length of each part becomes $L/2$.
Since $k \cdot L = k' \cdot (L/2)$, the new spring constant becomes $k' = 2k$.
The new frequency $f_2$ with the half spring is:
\[ f_2 = \frac{1}{2\pi}\sqrt{\frac{2k}{m}} \]
\[ f_2 = \sqrt{2} \left( \frac{1}{2\pi}\sqrt{\frac{k}{m}} \right) = \sqrt{2} f_1 \]
Taking the ratio of the two frequencies:
\[ \frac{f_2}{f_1} = \sqrt{2} \]
Step 4: Final Answer:
The ratio $f_2/f_1$ is $\sqrt{2}$.