AP EAMCET · PHYSICS · Oscillations
A spring is stretched by 0.2 m when a mass of 0.5 kg is suspended to it. The time period of the spring when 0.5 kg mass is replaced with a mass of 0.25 kg is suspended to it is
(Acceleration due to gravity \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) )
- A 0.628 s
- B 6.28 s
- C 62.8 s
- D 0.0628 s
Answer & Solution
Correct Answer
(A) 0.628 s
Step-by-step Solution
Detailed explanation
\( k = \frac{m_1g}{x_1} = \frac{0.5 \times 10}{0.2} = 25 \mathrm{~N/m} \) \( T = 2\pi\sqrt{\frac{m_2}{k}} = 2\pi\sqrt{\frac{0.25}{25}} \) \( T = 2\pi\sqrt{0.01} = 2\pi(0.1) = 0.2\pi \approx 0.628 \mathrm{~s} \)
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