Q90 / 160 · JIPMER 2007 · Physics

A particle executing simple harmonic motion of amplitude $5$ cm has maximum speed of $31.4 \mathrm{~cm} / \mathrm{s}$. The frequency of its oscillation is

  • A
    3 Hz
  • B
    2 Hz
  • C
    4 Hz
  • Correct answer
    D
    1 Hz

Solution

Maximum speed of a particle executing SHM is $u_{\max }=a \omega=a(2 \pi n)$ $\Rightarrow n=\frac{u_{\max }}{2 \pi a}$ Here, $u_{\max }=31.4 \mathrm{~cm} / \mathrm{s}, a=5 \mathrm{~cm}$ Substituting the given values, we have $n=\frac{31.4}{2 \times 3.14 \times 5}=1 \mathrm{~Hz}$

Hence, the correct answer is option: 'D'.

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