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

Options

A. `(2pi)/(omega)`
B. `(pi)/(omega)`
C. `(pi)/(2 omega)`
D. None of the above

Answer

A

Explanation

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 2 omega t," "T_(2)=(2pi)/(2 omega)=(pi)/(omega)`
and `y_(3)=sin 3 omega t," "T_(3)=(2pi)/(3 omega)`
`therefore" "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)`, the entire function repeats its motion.