Objects A and B are in harmonic motion modeled by yA = 8 sin(2t − π/3) and yB = 8 sin(2t − π/4) The phase difference between yA and yB is ....?

Answer:The phase difference between yA and yB is  [tex]\frac{-\pi }{12}[/tex] Step-by-step explanation:Given harmonic modeled as :yA = 8 sin(2t  - [tex]\frac{\pi }{3}[/tex])         And yB = 8 sin(2t  - [tex]\frac{\pi }{4}[/tex]) The function as written as :y = a sin(ωt  - Ф)  where  Ф is phase difference So , phase difference between yA   and  yB = ( Ф_1 - Ф_2 )Or phase difference between yA   and  yB = ( -  [tex]\frac{\pi }{3}[/tex] + [tex]\frac{\pi }{4}[/tex] )Or, phase difference between yA   and  yB = [tex](\frac{-4\pi +3\pi }{12})[/tex]I.e phase difference between yA   and  yB =[tex]\frac{-\pi }{12}[/tex]Hence The phase difference between yA and yB is [tex]\frac{-\pi }{12}[/tex]   Answer