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A partcle is performing simple harmonic motion along x- axis with amplitude `4cm` and time period `1.2sec` The minimum time period taken by the again is given byA. `0.6sec`B. `0.4 sec`C. `0.3 sec`D. `0.2 sec` |
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Answer» Correct Answer - B Time taken by period to move from `x = 0` (mean pisition) to `x = 4` (extermeposition) `= (T)/(4) = (1.2)/(4) = 0.3 sec` Let `t` be the time taken by the particle to move from `x = 0 to x= 2 cm` `y = a sin omega t rArr 2 = 4 sin"(2pi)/(T)t rArr (1)/(2) = sin"(2pi)/(1.2)t` `rArr (pi)/(6) = (2pi)/(1.2) t rArr = 0.1 sec` Hence time to move from `x = 2 to x= 4` will be equal to `0.3 - 0.1 = 0.2 sec` Hence total time to move from`x = 2 to x= 4`and back again ` 2 xx 0.2 = 0.4sec` |
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