MCQ
A particle moves such that its acceleration $a$ is given by $a = - bx$, where $x$ is the displacement from equilibrium position and b is a constant. The period of oscillation is
  • A
    $2\pi \sqrt b $
  • $\frac{{2\pi }}{{\sqrt b }}$
  • C
    $\frac{{2\pi }}{b}$
  • D
    $2\sqrt {\frac{\pi }{b}} $

Answer

Correct option: B.
$\frac{{2\pi }}{{\sqrt b }}$
b
(b) In the given case, $\frac{{Displacement}}{{Acceleration}} = \frac{1}{b}$
ime period $T = 2\pi \sqrt {\frac{{Displacement}}{{Acceleration}}}  = \frac{{2\pi }}{{\sqrt b }}$

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