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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` |
Answer» Correct Answer - 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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