Question

A spring stretches by \(2\,\text{mm}\) when it is loaded with a mass of \(200\,\text{g}\). From equilibrium position the mass is further pulled down by \(2\,\text{mm}\) and released. The frequency associated with the system and the maximum energy in the spring are _____ Hz and _____ J, respectively. (Take \(g=10\,\text{m/s}^2\)).

• A. \( \frac{5\sqrt{50}}{\pi} \) and \(8\times10^{-3}\)
• B. \( \frac{5\sqrt{50}}{\pi} \) and \(8\)
• C. \( \frac{10\sqrt{50}}{\pi} \) and \(2\times10^{-3}\)
• D. \( \frac{5\sqrt{50}}{\pi} \) and \(16\times10^{-3}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the frequency of oscillation of a spring-mass system and the maximum potential energy stored in the spring.
Step 2: Key Formula or Approach:
1. Spring constant \(k\) is found from static equilibrium: \(mg = k x_0\).
2. Frequency \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\).
3. Maximum potential energy in the spring is \(\frac{1}{2} k (x_{max})^2\), where \(x_{max}\) is the maximum extension from the natural length.
Step 3: Detailed Explanation:
Given: \(m = 200 \text{ g} = 0.2 \text{ kg}\), \(x_0 = 2 \text{ mm} = 0.002 \text{ m}\).
Static stretching: \(0.2 \times 10 = k \times 0.002 \Rightarrow k = \frac{2}{0.002} = 1000 \text{ N/m}\).
Frequency:
\[ f = \frac{1}{2\pi}\sqrt{\frac{1000}{0.2}} = \frac{1}{2\pi}\sqrt{5000} = \frac{1}{2\pi}(50\sqrt{2}) = \frac{1}{2\pi}(10\sqrt{50}) = \frac{5\sqrt{50}}{\pi} \text{ Hz} \]
The mass is pulled down an additional amplitude \(A = 2 \text{ mm}\).
Maximum extension from natural length \(x_{max} = x_0 + A = 2 \text{ mm} + 2 \text{ mm} = 4 \text{ mm} = 0.004 \text{ m}\).
Maximum energy in spring:
\[ U_{max} = \frac{1}{2} k (x_{max})^2 = \frac{1}{2} \times 1000 \times (0.004)^2 \]
\[ U_{max} = 500 \times 16 \times 10^{-6} = 8000 \times 10^{-6} = 8 \times 10^{-3} \text{ J} \]
Step 4: Final Answer:
The results match option (A).