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

Answer

C

Explanation

The particle starts from the mean position.
`therefore x=A "sin" omega t`
`therefore " for " t=1, x_(1)=A "sin" omega`
for `t=2, x_(2)=A "sin" 2omega`
`therefore (x_(1))/(x_(2))=("sin" omega)/("sin" 2 omega)=("sin" omega)/(2"sin" omega "cos" omega)`
`=(1)/(2 "cos" omega)=(1)/(2 "cos" ((2pi)/(T)))`
`therefore (x_(1))/(x_(2))=(1)/(2"cos" ((2pi)/(8)))=(1)/(2((1)/(sqrt(2))))=(1)/(sqrt(2))`
`therefore x_(2)=sqrt(2)x_(1)`
`therefore` The dis"tan"ce travelled in the second second
`D_(2)=sqrt(2)x_(1)-x_(1)=x_(1)(sqrt(2)-1) and D_(1)=x_(1)`
`therefore (D_(1))/(D_(2))=(1)/((sqrt(2)-1))`