A particle performing SHM has time period ` (2pi)/(sqrt(3))` and path lenght 4 cm. The displacement from mean position at which acceleration is equal to velocity is

Options

A. 0 cm
B. 0.5 cm
C. 1 cm
D. 1.5 cm

Answer

C

Explanation

Velocity ` v=omegasqrt(A^(2)-x^(2))` and acceleration ` =omega^(2)x`
Given, `omega sqrt(A^(2)-x^(2))=omega^(2)x`
or ` sqrt(A^(2)-x^(2))=omega^(2)x` ......(i)
Given , ` T=(2pi)/(sqrt(3))`
and ` omega=(2pi)/(T)=sqrt(3)`
Substituting the value of ` omega ` in Eq(i) , we get ` sqrt(A^(2)-x^(2))=sqrt(3)x`
`implies A =2 x `
As amplitude`=("path length")/(2)=2 cm `
`implies x=1 cm `