Question

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

Answer 1

The Time period of the function, y=sin omega t + sin 2omega t + sin 3omega t  is 2π/ω .

Given y = y1+ y2 + y3 , where

• y1 = sin ωt
• y2 = sin 2ωt
• y3 = sin 3ωt

Let time periods of y1, y2 and y3 be TP1 , TP2 and TP3

We know

• TP1 = 2π/ω

• TP2  = 2π/2ω = π/ω

• TP3 = 2π/3ω = 2/3 π/ω

Therefore,

• TP1 = 2TP2

• and TP1  = 3TP3

Therefore Time period of the function is TP1 = 2π/ω.

• Since it is the time for which sinωt to be periodic.
• Other functions sin2ωt and sin3ωt will go through 2 and 3 periods/oscillation, respectively, by the time sinωt goes through one period or one oscillation.