A particle performs a S.H.M. of period `2pi` sec and amplitude 2 cm. At what dis"tan"ce from the mean position, its velocity and acceleration are numerically equal ?

Options

A. `sqrt(2) cm`
B. `sqrt(3)cm`
C. `sqrt(6)cm`
D. `sqrt(5)cm`

Answer

A

Explanation

|Velocity|=|Acceleration |
`therefore omegasqrt(A^(2)-x^(2))=omega^(2)x therefore A^(2)-x^(2)=omega^(2)x^(2) and omega =(2pi)/(T)=(2pi)/(2pi)=1`
`therefore 4=x^(2)+x^(2) therefore x =sqrt(2)`