Simple Harmonic Motion - Phase Difference Question?

Question

I have been stuck on this question for quite a while now and I'm looking for some hints.

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?

Thanks in advance.

Answer 1

let x1 = a sin wt x2 = a sin (wt+p)

let at t = t1, x1=a/2 and x2 = -a/2 as given in data

a/2 = a sin wt1 therefore wt1 = pi/6 = 30 degrees -----------A

also wt1+p = pi+pi/6 = 210 degrees -----------------------------B

from A and B p, the phase difference = pi or 180 degrees

Answer 2

A simple harmonic motion is a vibratory motion in a line between two points. The speed at both the end points is 0, and maximum at the center. The motion is driven by an elastic restoring force proportional to the distance from the center. Acceleration being proportional to the force, hence it is always proportional to the distance from the center also and always is in the negative direction of the motion, as the force is a restoring force, always trying to draw the particle under motion to the center or the equilibrium position. If a particle undergoes a motion of uniform speed along a circle, its projection on a particular diameter undergoes a simple harmonic motion. Suppose the radius vector of the particle on the circle makes an angle theta = wt with a fixed radius and if the radius is of length A, ( the diameter being 2A) and if the fixed radius vector makes an angle Φ with fixed diameter, the radius vector of any point on the circle makes an angle wt + Φ with the fixed diameter and Acos(wt + Φ) is the projection of the revolving radius vector on the fixed diameter. (wt + Φ) is called the 'phase' of the moving point on the circle and Φ is called the 'phase angle'. Clearly, (wt + Φ) is the angle described by the moving point on the circle and A(wt + Φ) is the arc covered by it in time t, the distance of the projection on the diameter of the point moving on the circle is displaced by X = Acos(wt + Φ) from the center. Now for finding velocity of the projection point, differentiate it wrt X to get v =d X / d t= Awsin(wt + Φ) = Awcos(wt + Φ+90). This is described as ' the difference in phase between displacement and velocity of a particle executing SHM' is 90 degrees.A second time differentiation gives accelearation a = dv / dt = - Aw.wcos(wt + Φ), which is having phase difference of 180 degrees with the phase of displacement. Now a = dv / dt = - w.w.X, so, proportional to displacement and this is a simple harmonic motion by definition!.

Answer 3

If the particles passed going in the same direction, the phase difference would be arcsin(.5). Going in opposite directions, the phase difference is that plus 180º

The result is arcsin(.5)+(π/2)