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 is

Options

A. `1`
B. `1//3`
C. `1//4`
D. `1//2`

Answer

B

Explanation

`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)`