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InterviewSolution
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Find time period of the function, `y=sin omega t + sin 2omega t + sin 3omega t` |
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Answer» Correct Answer - A::B The given function can be written as, `y=y_(1) + y_(2) + y_(3)` Here, `y_(1)=sin omega t` , `T_(1)=(2pi)/(omega)` `y_(2)=sin 2omega t` , `T_(2)=(2pi)/(2omega)=(pi)/(omega)` and `y_(3)=sin 3omegat, T_(3)=(2pi)/(3omega)` `:. T_(1)=2T_(2)` and `T_(1)=3T_(3)` So, the time period of the given function is `T_(1)` or `(2pi)/(omega)`. Because in time `T=(2pi)/(omega)`, first function completes one oscillation, the second function two oscillation and the third three. |
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