Time period of a particle executing `SHM` is `8` sec. At `t=0` it is at the mean position. The ratio of the distance covered by the particle in the `1st` second to the `2nd` second is:

Options

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

Answer

D

Explanation

In 1st sec. `x=0` to `x=(A)/(sqrt(2))`
`d_(1)=(A)/(sqrt(2))`
In `2^(nd)` sec. `x=(A)/(sqrt(2))" to x = A"`
`d_(2)=A-(A)/(sqrt(2))implies(d_(1))/(d_(2))=sqrt(2)+1`