Two sine waves of the same period are described as π¦1 = 5π ππ (3π‘ β π/3) and π¦2 = 5π ππ (3π‘ β π/2), then the phase difference between y1 and y2 is (A) y1 leads y2 by 5π/6(B) y1 lags y2 by 5π/6 (C) y1 leads y2 by π/6
π¦1 = 5π ππ (3π‘ β π/3)
y1 = 5sin(3(t - Ο/9)
so y1 is Ο/9 to the right of y = 5sin(3t)
y2 = 5sin(3t - Ο/2)
y2 = 5sin(3(t - Ο/6) , moved Ο/6 to the right of y = 5sin 3t
so clearly y2 has moved to the right Ο/6 while y1 has moved Ο/9
difference = Ο/6 - Ο/9 is Ο/18 , which is not in the choices
graphing
y = 5sin(3(t - Ο/9) and
y = 5sin(3(t - Ο/6) on Desmos : www.desmos.com/calculator
verifies my answer.