The acceleration versus displacement graph of a particle performing SHM is shown in figure. The time period of oscillation of particle in second is

Options

A. `2pixxsqrt(2)`
B. `2pixxsqrt(3)`
C. `2pixx(2)^(1//4)`
D. `2pixx(3)^(1//4)`

Answer

D

Explanation

From the graph, `(a)/(x)=-tan30^(@)=-(1)/(sqrt(3))`
or`a=-(x)/(sqrt(3))`
In SHM, acceleration, `a=-omega^(2)x`, therefore,
`omega^(2)=(1)/(sqrt(3))`
o r `omega=((1)/(sqrt(3)))^(1//2)=((1)/((3)^(1//4))) or (2pi)/(T)=((1)/(3))^(1//4)`
or `T=2pi(3)^(1//4)`