Two sine waves of the same period are described as 𝑦1 = 5𝑠𝑖𝑛 (3𝑡 − 𝜋/3) and 𝑦2 = 5𝑠𝑖𝑛 (3𝑡 − 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

Answer 1

𝑦1 = 5𝑠𝑖𝑛 (3𝑡 − 𝜋/3)
y1 = 5sin(3(t - π/9)
so y1 is π/9 to the right of y = 5sin(3t)

y2 = 5sin(3t - π/2)
y2 = 5sin(3(t - π/6) , moved π/6 to the right of y = 5sin 3t

so clearly y2 has moved to the right π/6 while y1 has moved π/9

difference = π/6 - π/9 is π/18 , which is not in the choices

graphing
y = 5sin(3(t - π/9) and
y = 5sin(3(t - π/6) on Desmos : www.desmos.com/calculator

verifies my answer.