AP EAMCET · PHYSICS · Oscillations
When a body of mass \(1.0 \mathrm{~kg}\) is suspended from a certain light spring hanging vertically, its length increases by \(5 \mathrm{~cm}\). By suspending \(2.0 \mathrm{~kg}\) block to the spring and if the block is pulled through \(10 \mathrm{~cm}\) and released, the maximum velocity in it in \(\mathrm{m} / \mathrm{s}\) is : (Acceleration due to gravity \(=10 \mathrm{~m} / \mathrm{s}^2\) )
- A 0.5
- B 1
- C 2
- D 4
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
\(m_1=1 \mathrm{~kg}\), extension \(l_1=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}\) \(\therefore \quad m_1 g=k l_1\) \(k=\) force constant of the spring \(k=\frac{m_1 g}{l_1}=\frac{1 \times 10}{5 \times 10^{-2}}=200 \mathrm{~N} / \mathrm{m}\) Time period of the block of mass…
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