AP EAMCET · PHYSICS · Oscillations
A body of mass 1 kg is suspended from a spring of force constant \(600 \mathrm{~N} \mathrm{~m}^{-1}\). Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of \(3 \mathrm{~m} \mathrm{~s}^{-1}\) and embedded in it. The amplitude of motion is
- A 5 cm
- B 15 cm
- C 10 cm
- D 8 cm
Answer & Solution
Correct Answer
(A) 5 cm
Step-by-step Solution
Detailed explanation
\(M = m_1 + m_2 = 1 + 0.5 = 1.5 \text{ kg}\) \(V = \frac{m_2 v_2}{M} = \frac{0.5 \times 3}{1.5} = 1 \text{ m s}^{-1}\) \(x_0 = -\frac{m_2 g}{k} = -\frac{0.5 \times 10}{600} = -\frac{5}{600} = -\frac{1}{120} \text{ m}\) (assuming \(g=10 \text{ m s}^{-2}\))…
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