Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. each particle has a period of 1.5 s, but they differ in phase by π/6 rad.

(a) how far apart are they 0.45 s after the lagging particle leaves one end of the path?
(b) are they then moving in the same direction, toward each other, or away from each other?

a) the particles are 0.217 m apart

b) the particles are moving in the same direction.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

    = -0.155 A + 0.372 A

    = 0.217 A

since A = 1 m

Thus,

Δx  = 0.217 m

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

   = (-πA / T) sin(2πt / T)

   = -(π(1) / 1.5) sin(2π(0.45) / 1.5)

   = -1.99

and,

v₂ = dx₂ / dt

   = (-πA / T) sin((2πt / T) + π/6)

   = -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

   = -1.40

Since both v₁ and v₂ are negative, this shows that the particles are moving in the same direction.

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