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Question: A particle moves such that its acceleration *a* is given by \(a = - b x\). Where *x* is the displace...

A particle moves such that its acceleration a is given by a=−bxa = - b x. Where x is the displacement from equilibrium position and b is a constant. The period of oscillation is

A

2πb2 \pi \sqrt { b }

B

2πb\frac { 2 \pi } { \sqrt { b } }

C

2πb\frac { 2 \pi } { b }

D

2πb2 \sqrt { \frac { \pi } { b } }

Answer

2πb\frac { 2 \pi } { \sqrt { b } }

Explanation

Solution

We know that Acceleration =−ω2- \omega ^ { 2 } (displacement) and

a=−bxa = - b x (given in the problem)

Comparing above two equation ω2=b\omega ^ { 2 } = b⇒ ω=b\omega = \sqrt { b }

∴ Time period T=2πω=2πbT = \frac { 2 \pi } { \omega } = \frac { 2 \pi } { \sqrt { b } }