InterviewSolution
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InterviewSolution
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A body has a time period `T_(1)` under the action of one force and `T_(2)` under the action of another force, the square of the time period when both the forces are acting in the same direction isA. `T_(1)^(2)T_(2)^(2)`B. `T_(1)^(2)//T_(2)^(2)`C. `T_(1)^(2) + T_(2)^(2)`D. `T_(1)^(2)T_(2)^(2)//(T_(1)^(2)+T_(2)^(2))` |
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Answer» Correct Answer - D As `F_(1) = (m 4 pi^(2)a)/(pi^(2)) and F_(2) = (m 4 pi^(2)a)/(T_(2)^(2))` Net force, `F = F_(1) + F_(2) = 4 pi^(2) ma((1)/(T_(1)^(2))+(1)/(T_(2)^(2))) or (4 pi^(2)ma)/(T_(2))=4pi^(2)ma((1)/(T_(1)^(2))+(1)/(T_(2)^(2)))` or `(1)/(T^(2))=(1)/(T_(1)^(2))+(1)/(T_(2)^(2))" "rArr" "T^(2) = (T_(1)^(2)T_(2)^(2))/(T_(1)^(2)+T_(2)^(2))` |
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