The period of a particle in SHM is `8s`. At `t=0` it is at the mean position. The ratio of the distances travled by it in the first and the second second is

Options

A. `(sqrt(2) - 1)/(sqrt(2))`
B. `(1)/(sqrt(2))`
C. `(1)/(sqrt(2) - 1)`
D. `[sqrt(2) - 1]`

Answer

C

Explanation

`:' x = sin omegat`
`:. x_(1) = A sin omega` & `x_(1) + x_(2) = Asin (2omega)`
`omega = (2pi)/(T) = (2pi)/(8) = (pi)/(4) rArr x_(1) = (A)/(sqrt(2))` & `x_(1) + x_(2) = A`
`rArr x_(2) = A - (A)/(sqrt(2))`, Therefore `(x_(1))/(x_(2)) = (1)/(sqrt(2)- 1)`
OR
Suppose amplitude be A and distance traveled in `1 sec` be `x_(1)` and in `2sec` be `x_(2)`.
`phi = omegat = (2pi)/(8)1 = (pi)/(4)`
Therefore `0 = (pi)/(4)`
`costheta = cos"(pi)/(4) = (x_(1))/(A)`

`rArr x_(1) = (A)/(sqrt(2))` & `x_(1) + x_(2) = A`
`rArr x_(2) = A((sqrt(2) - 1)/(sqrt(2)))`. Therefore `(x_(1))/(x_(2)) = (1)/(sqrt(2) - 1)`