A particle is moving in a circel of radius `R = 1m` with constant speed `v =4 m//s`. The ratio of the displacement to acceleration of the foot of the perpendicular drawn from the particle onto the diameter of the circel is

Options

A. `1//16(second)^(2)`
B. `1//2(second)^(2)`
C. `2(second)^(2)`
D. `16(second)^(2)`

Answer

A

Explanation

As the particles moves in circular path with angular velocity `omega`, the foot of its perpendicular on the dimeter executes SHM of same angular frequency `omega`.
`omega=(V)/(R) Rightarrow ("displacement it foot")/("acceleration of foot")=(x)/(omega^(2)x)=(1)(omega^(2))`
`(x=Asintheta=Asinomegat)`
`x=Asintheta)`
`Asinomegat`