AP EAMCET · PHYSICS · Oscillations
A body of mass 1 kg is attached to the lower end of a vertically suspended spring of force constant \(600 \mathrm{~N} \mathrm{~m}^{-1}\). If another body of mass 0.5 kg moving vertically upward hits the suspended body with a velocity \(3 \mathrm{~m} \mathrm{~s}^{-1}\) and embedded in it, then the frequency of the oscillation is
- A \(\frac{5}{\pi} \mathrm{~Hz}\)
- B \(\frac{10}{\pi} \mathrm{~Hz}\)
- C \(\frac{\pi}{5} \mathrm{~Hz}\)
- D \(\pi \mathrm{Hz}\)
Answer & Solution
Correct Answer
(B) \(\frac{10}{\pi} \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
\(M = m_1 + m_2 = 1 \mathrm{~kg} + 0.5 \mathrm{~kg} = 1.5 \mathrm{~kg}\) \(f = \frac{1}{2\pi}\sqrt{\frac{k}{M}} = \frac{1}{2\pi}\sqrt{\frac{600 \mathrm{~N} \mathrm{~m}^{-1}}{1.5 \mathrm{~kg}}} = \frac{1}{2\pi}\sqrt{400} = \frac{1}{2\pi}(20) = \frac{10}{\pi} \mathrm{~Hz}\)
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