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Phase Difference

A particle mo...

Question

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

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B

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C

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D

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Solution

The correct option is B

In the given case , \( \frac {Displacement }{Acceleration } = \frac 1b )

$∴ T i m e p e r i o d T = 2 π \sqrt{\frac{D i s p l a c e m e n t}{A c c e l e r a t i o n}} = \frac{2 π}{\sqrt{b}}$

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