A particle moves in such a way that its acceleration a= - bx, where x is its displacement from the mean position and b is a constant. The period of its oscillation is

Options

A. `2 pi b`
B. `(2pi)/(b)`
C. `(2pi)/(sqrt(b))`
D. `2pisqrt(b)`

Answer

C

Explanation

a =- bx or a +bx=0 comparing it with
`(a^(2)x)/(dt^(2))+omega^(2)x=0`
We get `omega^(2)=b " " therefore (4pi^(2))/(T^(2))=b " " therefore T=(2pi)/(sqrt(b))`