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Find time period of the function, y=sin ...

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

Text Solution

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