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. When a spring is cut, its spring constant changes inversely with its length. We need to determine the new spring constant and then evaluate the new frequency.
Step 2: Key Formula or Approach:
1. Frequency of oscillation: $\nu = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
2. Spring constant is inversely proportional to length: $k \propto \frac{1}{L} \implies k \cdot L = \text{constant}$.
Step 3: Detailed Explanation:
Let the initial length of the spring be $L$ and its spring constant be $k$.
The initial frequency is given by:
$\nu_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$.
When the spring is cut to half its length, the new length is $L' = L/2$.
Since $k \times L = k' \times L'$, we have:
$k \cdot L = k' \cdot \left(\frac{L}{2}\right) \implies k' = 2k$.
The new spring constant is twice the original.
The new frequency $\nu_2$ with the same mass $m$ is:
$\nu_2 = \frac{1}{2\pi} \sqrt{\frac{k'}{m}} = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}$.
Rewrite $\nu_2$ in terms of $\nu_1$:
$\nu_2 = \sqrt{2} \left( \frac{1}{2\pi} \sqrt{\frac{k}{m}} \right) = \sqrt{2} \nu_1$.
Therefore, the ratio is:
$\frac{\nu_2}{\nu_1} = \sqrt{2}$.
Step 4: Final Answer:
The value of the ratio is $\sqrt{2}$.