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A particle moves such that its accelerat...

A particle moves such that its acceleration a is given by a = -bx , where x is the displacement from equilibrium positionand is a constant. The period of oscillation is

A

`2pisqrtb`

B

`(2pi)/(sqrtb)`

C

`(2pi)/(b)`

D

`2sqrt((pi)/(b))`

Text Solution

Verified by Experts

The correct Answer is:
B
Acceleration (a)=-bx
Since, for SHM
`a=-omega^(2)x`
`therefore omega^(2)=b implies omega=sqrtb`
`therefore T=(2pi)/(omega)=(2pi)/(sqrtb)`.
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