Find time period of the function, `y=sin omega t + sin 2omega t + sin 3omega t`
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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.