A particle oscillates simple harmonically along a straight line with period 8 seconds and amplitude `8 sqrt(2)`m. It starts from the mean position, then the ratio of the distances travelled by it in the second second and first second of its motion is

Options

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

Answer

C

Explanation

`8 sqrt(2)m and T=8 s and omega =(2pi)/(8)=(pi)/(4)`
The particle starts from the mean position,
`therefore x= A "sin" omega t`
`therefore` The dis"tan"ce travelled by the particle in one second is
`therefore x_(1)=8sqrt(2) "sin" ((pi)/(4)xx1)`
`=8sqrt(2)xx(1)/(sqrt(2))=8 m =s_(1)`
and the dis"tan"ce travelled by the particle in 2 s is
`x_(2)=8sqrt(2) "sin" ((pi)/(4)xx2)=8sqrt(2)m`
`therefore` Dis"tan"ce travelled in second second `(s_(2))`
`=x_(2)-x_(1)=8sqrt(2)-8=8(sqrt(2)-1)m`
`therefore (s_(2))/(s_(1))=(8(sqrt(2)-1))/(8)=sqrt(2)-1`