A particle is executing SHM with and frequency of `(1)/(8)` Hz. If it starts from the mean position at time t = 0, the ratio of distances convered by it in 1st and 2nd seconds is

Options

A. 1
B. `1//(sqrt(2)-1)`
C. `1//(sqrt(3)-1)`
D. `sqrt(2)-1`

Answer

B

Explanation

`n=(1)/(8)HzimpliesT=8s`
Mean to extreme t = 2s. The distance covered by it in 1 second is `y_(1)` and in the II second `=A-y_(1)`.
`(y_(1))/(A-y_(1))=(Asin.(2pi)/(8)xx1)/(A-Asin.(2pi)/(8)xx1)=(A//sqrt(2))/(A-A//sqrt(2))=(1)/(sqrt(2)-1)`