A particle doing simple harmoonic motion amplitude `= 4 cm` time peiod `= 12sec` The ratio between time taken by it in going from its mean position to `2cm ` and from `2cm` to exterme position isA. `1`B. `1//3`C. `1//4`D. `1//2`

A particle doing simple harmoonic motion amplitude `= 4 cm` time peiod

1.	A particle doing simple harmoonic motion amplitude `= 4 cm` time peiod `= 12sec` The ratio between time taken by it in going from its mean position to `2cm ` and from `2cm` to exterme position isA. `1`B. `1//3`C. `1//4`D. `1//2`
Answer» Correct Answer - B `omega = (2pi)/(T) = (2pi)/(12) = (pi)/(6)(rod)/(sec)` (for `y = 2 cm) 2 = 4(sin"(pi)/(6)t_(1))` By solving ` t_(1) = 1` sec(For `y = 4 cm`) `t_(2) = 3 sec` So time taken by particle from `2cm` to extreme position is `t_(2) - t_(1) = 2 sec` Hence required ratio will be `(1)/(2)`

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A particle doing simple harmoonic motion amplitude `= 4 cm` time peiod `= 12sec` The ratio between time taken by it in going from its mean position to `2cm ` and from `2cm` to exterme position isA. `1`B. `1//3`C. `1//4`D. `1//2`

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