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2 April, 22:52

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?

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  1. 2 April, 23:21
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    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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