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InterviewSolution
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A body is in `SHM` with period `T` when oscillated from a freely suspended spring. If this spring is cut in two parts of length ratio `1 : 3 &` again oscillated from the two the two parts separatedly, then the periods are `T_(1) & T_(2)` then find `T_(1)//T_(2)`. |
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Answer» As we know know, spring constant is inversely is inversely proportional to its natural length . Therefore. `k = ( C)/(l) : k_(1) = ( C)/(l_(1)) and k_(2) = ( C)/(l_(2))` `:. (k_(1))/(k_(2)) = (l_(2))/(l_(1)) = 3` `:. k_(1) = 3 k_(2)` But `(1)/(k) = (1)/(k_(1)) + (1)/(k_(2)) or (1)/(k) = (1)/(3 k_(2)) + (1)/(k_(2)) = (1 + 3)/(3 k_(2))` `k = (3 k_(2))/(4)` `:. k_(2) = (4 k)/(3)` `:. k_(1) = 3 k_(2) - 4k` But `T = 2 pi sqrt((m)/(k))` : `T_(1) = 2 pi sqrt((m)/(k_(1)))` and `T_(2) = 2 pi sqrt((m)/(k_(2)))` `:. (T_(1))/(T_(2)) = sqrt((k_(2))/(k_(1))) = sqrt((4k)/(3 xx 4k))` `(T_(1))/(T_(2)) = sqrt((1)/(3)) = 1: sqrt3` |
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