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. `1/2`
B. `1/sqrt2`
C. `sqrt2`
D. `(1)/(sqrt2-1)`

Answer

D

Explanation

`T=8s`, `T/4=25`
The particle moves in a straight line without change in direction so displacement and distance are same in first `2s`
`t=1s`, `x_1=Asinomegat`
`=Asinomega=Asin((2pi)/(T))=Asin((2pi)/(8))=A/sqrt2`
`t=2s`, `x_2=A`
`x_2-x_1=A(1-sinomega)=A(1-sin((2pi)/(T)))`
`A(1-sin((2pi)/(8)))=A(1-(1)/(sqrt2))`
`(x_1)/(x_2-x_1)=(A//sqrt2)/(A(1-1/sqrt2))=(1)/(sqrt2-1)`