Question

The frequency of oscillation of a mass \(m\) suspended by a spring is \(v_1\). If the length of the spring is cut to half, the same mass oscillates with frequency \(v_2\). The value of \(\frac{v_2}{v_1}\) is ________.

• A. 1
• B. 2
• C. \(\sqrt{2} \)
• D. \(\sqrt{3} \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The frequency of a spring-mass system depends on the spring constant $k$ and the mass $m$. When a spring is cut, its spring constant changes inversely with its length.
Step 2: Key Formula or Approach:
1. Frequency $v = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$.
2. Spring constant relation: $k \propto \frac{1}{L} \implies k \cdot L = \text{constant}$.
Step 3: Detailed Explanation:
Let the initial length be $L_1$ and initial spring constant be $k_1$. The initial frequency is $v_1 = \frac{1}{2\pi}\sqrt{\frac{k_1}{m}}$. When the spring is cut to half ($L_2 = L_1/2$), the new spring constant $k_2$ is: \[ k_2 \cdot \frac{L_1}{2} = k_1 \cdot L_1 \implies k_2 = 2k_1 \] The new frequency $v_2$ is: \[ v_2 = \frac{1}{2\pi}\sqrt{\frac{2k_1}{m}} = \sqrt{2} \left( \frac{1}{2\pi}\sqrt{\frac{k_1}{m}} \right) \] \[ v_2 = \sqrt{2} \cdot v_1 \] Therefore, $\frac{v_2}{v_1} = \sqrt{2}$.
Step 4: Final Answer:
The value of the ratio is $\sqrt{2}$.