Two particle undergoes `SHM` along parallel line with the same time period `(T)` and equal amplitude At a particular instant , one particle is at its extereme position while the other is at its mean position .They move in the same direction .They will cross each other after a further time

Options

A. `7//8`
B. `37//8`
C. `2//6`
D. `4//13`

Answer

B

Explanation

The problem is may to move with the help of phase diagram.Find we draw initial position of both the particle on the phase as shown in figure

From above figure difference between both the particle is `pi//2` They will cross other when their propertion from the circle in the horizontal diamater meat at the point
Let after time t both with each at `P'Q'` point haxing phese diffarence `pi2` as shown in figure

Both will ment at `- A sqrt(2)`
When they angle displacement at `P` is
`theta = (pi)/(2) + (pi)/(4) = (3pi)/(4)`
So they will ment line `t = (3pi)/(4 xx omega)`
`t = (3pi)/(4 xx 2pi) xx T = (3T)/(8)sec`