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JIPMER 2007
Exam
JIPMER
Year
2007
Subject
Physics
Chapter
Oscillations

JIPMER 2007 · Physics · Oscillations

A particle executing simple harmonic motion of amplitude 5 cm has...

A particle executing simple harmonic motion of amplitude 55 cm has maximum speed of 31.4 cm/s31.4 \mathrm{~cm} / \mathrm{s}. The frequency of its oscillation is

Options

  1. A3 Hz
  2. B2 Hz
  3. C4 Hz
  4. D1 Hz

Correct answer

D. 1 Hz

Step-by-step solution

Maximum speed of a particle executing SHM is umax⁡=aω=a(2πn)u_{\max }=a \omega=a(2 \pi n) ⇒n=umax⁡2πa\Rightarrow n=\frac{u_{\max }}{2 \pi a} Here, umax⁡=31.4 cm/s,a=5 cmu_{\max }=31.4 \mathrm{~cm} / \mathrm{s}, a=5 \mathrm{~cm} Substituting the given values, we have n=31.42×3.14×5=1 Hzn=\frac{31.4}{2 \times 3.14 \times 5}=1 \mathrm{~Hz}
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