Question

Two particles oscillate in simple harmonic motion along a common straight-line segment of length

a. each particle has a period of 1.5 s, but they differ in phase by pi/6 rad. (a) how far apart are they (in terms of

a.0.50 s after the lagging particle leaves one end of the path

Answer 1

Assuming the two functions are sine functions.
From the problem:
T = 1.5
So
b = 2π/T
b = 2π/1.5
b = 4π/3

The first sine function has an equation
y1 = sin 4π/3x
and since they have a difference in phase of π/6, the other sine function has an equation
y2 = sin 4π/3x + π/6

Their difference is simply π/6.
If we substitute 0.5 s to each equation,
y1 = 2π/3
y2 = 5π/6