MCQ
Values of the acceleration $A$ of a particle moving in simple harmonic motion as a function of its displacement $x$ are given in the table below. The period of the motion is

$A (mm \,\,s^{-2}$)

 $16$

    $8$

$0$

$- 8$

$- 16$

$x\;(mm)$

$- 4$

$- 2$

$0$

  $2$

   $4$

  • A
    $\frac{1}{\pi }s$
  • B
    $\frac{2}{\pi }s$
  • C
    $\frac{\pi }{2}s$
  • $\pi \,s$

Answer

Correct option: D.
$\pi \,s$
d
(d) $|A| = \omega^2x$ ==> $\frac{{|A|}}{x} = {\omega ^2}$

From the given value $\frac{{|A|}}{x} = {\omega ^2} = 4$ ==> $\omega = 2.$

Also $\omega = \frac{{2\pi }}{T} \Rightarrow 2 = \frac{{2\pi }}{T}\, \Rightarrow T = \pi \,sec$

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