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

Options

A. `sqrt2:1`
B. `1:(sqrt2-1)`
C. `(sqrt2+1):sqrt2`
D. `(sqrt2-1):1`

Answer

B

Explanation

`x=asinomegat` with `omega=(2pi)/(T)=(2pi)/(8)=(pi)/(4)(rad)/(s)`
`x=asin(pit)/(4)`
In 2s (which is equal to `(T)/(4)`), one amplitude will be convered. In 1st second
`x=asin((pi)/(4))=(a)/(sqrt2)`
Required ratio`=((a)/(sqrt2))/(a-(a)/(sqrt2))=(1)/(sqrt2-1)`