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InterviewSolution
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A body has a time period (4/5) s under the action of one force and (3/5) s under the action of another force. Then time period when both the forces are acting in the same direction simultaneously will beA. `4//5 s`B. `5//4 s`C. `1 s`D. `12//25 s` |
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Answer» Correct Answer - D `F=F_(1)+F_(2)` `T^(2)prop(1)/(F) therefore Fprop(1)/(T^(2))` `T=(T_(1)T_(2))/(sqrt(T_(1)^(2)+T_(2)^(2)))=((4)/(5)xx(3)/(5))/(sqrt((16)/(25)+(9)/(25)))=(12//25)/(1)` `therefore T=0.48` |
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