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

Answer

A

Explanation

Since the particle is at the mean position at `t=0`, so its displacement x any time t is
`x=Asin(omegat)=Asin((2pi)/(T)t)=Asin((pi)/(4)t)(because T=8s)`
The distance travelled in the `I^(st)` second is
`x_(1)=Asin((pi)/(4))=(A)/(sqrt(2))`
and that travelled in the `2^(nd)` second is
`x_(2)=Asin((pi)/(2))-Asin((pi)/(4))=A-(A)/(sqrt(2))=A(1-(1)/(sqrt(2)))`
`therefore (x_(1))/(x_(2))=((A)/(sqrt(2)))/(A(1-(1)/(sqrt(2))))=(1)/(sqrt(2)-1)`