Answer:
a) the particles are 0.217 m apart
b) the particles are moving in the same direction.
Step-by-step 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.