The function of time representing a simple harmonic motion with a period of π ω is :

Step-by-step solution

Time period T = 2 π ω ' π ω = 2 π ω ' ω ' = 2 ω → Angular frequency of SHM Option ( 3 ) sin 2 ω t = 1 2 2 sin 2 ω t = 1 2 1 - cos 2 ω t Angular frequency of 1 2 - 1 2 cos 2 ω t is 2 ω Option ( 4 ) Angular frequency of SHM 3 cos π 4 - 2 ω t is 2 ω . So option ( 3 )   &   ( 4 ) both have angular frequency 2 ω but option ( 4 ) is direct answer.