SHM - Finding Distance and Direction of Oscillating Particles

[LandOfStandar]

SHM - Simple Harmonic Motion Question

Homework Statement

Two particles oscillate in SHM along a common straight-line segment of length A. Each particle has a period of 1.5s, but they differ in phase by π/6 rad.
a. How far apart are they 9in terms of A) 0.50s after the lagging particle leaves one end of the path?
b. Are they then moving in the same direction, towards each other, or away from each other?

Homework Equations

X = A cos(ωt + θ)
T = 2π/ω

The Attempt at a Solution

T = 2π/ω Tω = 2π ω = 2π/T
X = A cos(ωt + θ) = A cos [(2π/T)t + θ] =
A cos [(2π/1.5)t + 0] and A cos [(2π/1.5)t + π/6]
A cos [(2π/1.5)t + 0] - A cos [(2π/1.5)t + π/6], t equals .5s
(-.50)A - (-.87)A = .37A, but the answer in the book is .18A.
What did I do wrong?

 

[LandOfStandar]

Bumping up so someone could tell me what the think.

 

[LandOfStandar]

anyone

 

[braska '92]

Just to give the correct answer. Your passages were all good except for the first one. The equations describing the motion of the particles are: (A/2)cos(wt + H) and (A/2)cos(wt) and not just A, because the segment is long A and so (knowing that the cos goes from -1 to + 1) a partile moves from A/2 to -A/2. Substituting A/2 the computation should be right!

 

[rwisz]

It seems that you have made a small error in your calculation for the distance between the two particles. The correct answer is indeed 0.18A, which can be found by subtracting the two positions of the particles at t=0.5s.

Let's take a closer look at your calculation. You correctly found the expression for the position of each particle, but the phase difference between them should be π/3 rad, not π/6 rad. This is because the lagging particle has a phase of 0 at t=0, while the leading particle has a phase of π/3 at t=0.

So the correct expression for the position of the particles would be A cos[(2π/1.5)t + 0] and A cos[(2π/1.5)t + π/3]. Plugging in t=0.5s, we get -0.87A and -0.50A, respectively. The difference between these two positions is indeed 0.18A.

As for the direction of motion, since the particles have the same period, they will always be moving in opposite directions. At t=0.5s, the lagging particle will be moving towards the end of the path, while the leading particle will be moving away from it.

I hope this helps clarify your doubts and leads you to the correct solution. Keep up the good work!