MCQ
A particle is excuting a simple harmonic motion. Its maximum acceleration is $\alpha $ and maximum velocity is $\beta $. Then its frequency of vibration will be
  • $\frac{\alpha }{{2\pi \beta }}$
  • B
    $\frac{{{\alpha ^2}}}{{{\beta ^2}}}$
  • C
    $\frac{\beta }{\alpha }$
  • D
    $\frac{\alpha }{{{\beta ^2}}}$

Answer

Correct option: A.
$\frac{\alpha }{{2\pi \beta }}$
a
$a_{\max }=\alpha \Rightarrow \omega^{2} A=\alpha$           $...(1)$

$\mathrm{V}_{\max }=\beta \Rightarrow \omega \mathrm{A}=\beta$          $...(2)$

Dividing

$\omega=\frac{\alpha}{\beta} \Rightarrow n=\frac{\alpha}{2 \pi \beta}$

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