Question

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

Answer 1

Correct Answer - C
Given, `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 distance travelled by the particle in one second is
`therefore" "x_(1)=8 sqrt(2)sin((pi)/(4)xx1)`
`= 8 sqrt(2)xx(1)/(sqrt(2))=8m = s_(1)`
and the distance travelled by the particle in 2s is
`x_(2)=8 sqrt(2)sin((pi)/(4)xx2)=8 sqrt(2)m`
`therefore` Distance travelled in second second `(s_(2))`
`=x_(2)-x_(1)=8 sqrt(2)-8 =(sqrt(2)-1)m`
`therefore" "(s_(2))/(s_(1))=(8 sqrt(2)-1)/(8)=sqrt(2)-1`