TS EAMCET · Physics · Oscillations
The force ( F in newton) acting on a particle of mass 90 g executing simple harmonic motion is given by \(\mathrm{F}+0.04 \pi^2 y=0\), where \(y\) is displacement of the particle in meter. If the amplitude of the particle is \(\frac{6}{\pi} \mathrm{~m}\), then the maximum velocity of the particle is
- A \(6 \mathrm{~ms}^{-1}\)
- B \(2 \mathrm{~ms}^{-1}\)
- C \(8 \mathrm{~ms}^{-1}\)
- D \(4 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(4 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\(F = -0.04 \pi^2 y \implies k = 0.04 \pi^2\) \(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{0.04 \pi^2}{0.09}} = \sqrt{\frac{4 \pi^2}{9}} = \frac{2\pi}{3} \mathrm{~rad/s}\) \(v_{max} = A\omega = \frac{6}{\pi} \times \frac{2\pi}{3} = 4 \mathrm{~ms}^{-1}\)
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