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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-
Minimum and maximum distance between A and B during the motion is-

A

`sqrt(5) cm` and `sqrt(61) cm`

B

`3 cm` and `7 cm`

C

`1 cm` and `5 cm`

D

`9 cm` and `16 cm`

Text Solution

Verified by Experts

The correct Answer is:
B
Distance between A & B
`= sqrt(x^(2) + y^(2))`
`= sqrt((3-2 cos 2pit)^(2) + (4-2 sin 2pit)^(2))`,
`= sqrt(9+4cos^(2)2pit - 12 cos2pit+16+4 sin^(2)2pit - 16sin2pit)`
`= sqrt(29-20(3/5 cos 2pit + 4/5 sin 2pit))`
`= sqrt(29-20sin(2pit + 37^(@)))`
Maximum distance
`= sqrt(29 + 20) = sqrt(49) = 7 cm`
Minimum distance
`= sqrt(29-20) = sqrt(9) = 3 cm`
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