Two particles `A` and `B` performing `SHM` along `x` and `y-`axis respectively with equal amplitude and frequency of `2 cm` and `1 Hz` respectively. Equilibrium positions of the particles `A` and `B` are at the co-ordinates `(3, 0)` and `(0, 4)` respectively. At `t = 0, B` is at its equilibrium positions and moving towards the origin, while `A` is nearest to the origin and moving away from the origin. If the maximum and minimum distances between `A` and `B` is `s_(1)` and `s_(2)` then find `s_(1) + s_(2)` (in `cm`).

Answer

10

Explanation

At `t = 0` Particle `2` is at point `V` and mocing towards origin so displacement
`Y = 4 - A sin omegat`
`Y = 4 - 2 sin omegat`
and displacement of particle `1` is
`X = 3 - A cos omegat `
`X = 3 - 2 cos omegat`
So distance between them `= sqrt(X^(2) + Y^(2))`
`s^(2) = 29 - (16 sin omegat + 12 cos omegat) = 29 - 4 (4 sinomegat + 3 cosomegat)`
`rArr 29 - 20 (sinomegat + 37^(@))`
So `s_(max)^(2) = 49 rArr s_(max) = 7 cm = S_(1)`
`s_(min)^(2) = 9 rArr s_(min) = 3cm = S_(2)`
`:. S_(1) + S_(2) = 10 cm`