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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

  1. A \(6 \mathrm{~ms}^{-1}\)
  2. B \(2 \mathrm{~ms}^{-1}\)
  3. C \(8 \mathrm{~ms}^{-1}\)
  4. D \(4 \mathrm{~ms}^{-1}\)
Verified Solution

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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