Two particles A and B are performing SHM...

Two particles A and B are 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 coordinates `[3 cm, 0]` and `(0, 4 cm)` respectively. At `t = 0 ,B` is at its equilibrium position and moving towards the origin, while A is nearest to the origin and moving away from the origin-
Equation of motion of particle A can be written as-

`x = (2 cm) cos 2pit`

`x = (3 cm) - (2 cm) cos 2 pit`

`x = (2 cm) sin 2 pit`

`x = (3 cm) - (2 cm) sin 2pit`

Verified by Experts

The correct Answer is:

As A is at its negative extreme at `t = 0`
so `x - 3 = 2 sin (2pit + 3pi//2) rArr x = 3-2 cos (2pit)`

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The displacement of a particle performing SHM x= 3sin 2t+4 cos 2t . The amplitude and maximum velocity of a particle respectively are………….

`5,10`

`3,2`

`4,2`

`3,4`

`144 pi^(2) m"/"sec^(2)`

`144 m"/"sec^(2)`

`(144)/(pi^2) m"/"sec^(2)`

`288 pi^(2) m"/"sec^(2)`

`2pi sqrt((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))`

`2pi sqrt((-x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)-v_(2)^(2)))`

`2pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`

`2pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`