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. 1 cm
B. 2 cm
C. 3 cm
D. 4 cm

Answer

A

Explanation

`omega sqrt(A^(2)-x^(2))= omega^(2) x therefore sqrt(A^(2)-x^(2)) = omega x`
`therefore A^(2)-x^(2)=omega^(2)x^(2) therefore A^(2)-x^(2)(omega^(2) +1)`
But `omega^(2)=(4pi^(2))/(T^(2))=(4pi^(2)xx3)/(4pi^(2))=3`
`therefore " Form " (1), 4-x^(2)(3+1)=4x^(2) therefore x=1 cm`