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Vibrating particle velocity, acceleration and path

Short answer

At t=1 s, v_x = -314 mm/s, v_y = 0, so velocity is 0.314 m/s along -x; a_x = 0, a_y = -987 mm/s², acceleration 0.987 m/s² along -y. The path is a parabola y = 25 - x²/200 (mm).

The motion of a vibrating particle is defined by the equations x = 100 sin πt and y = 25 cos 2πt where x and y are expressed in mm and t in sec. a) Determine the velocity and acceleration when t = 1 sec. b) Find the nature of path of the particle.

Velocity and acceleration (mm, s)

  • x = 100 sin πt, so vₓ = 100π cos πt, aₓ = −100π² sin πt
  • y = 25 cos 2πt, so v_y = −50π sin 2πt, a_y = −100π² cos 2πt

(a) At t = 1 s

  • vₓ = 100π cos π = −314.16 mm/s
  • v_y = −50π sin 2π = 0
  • v = 314.16 mm/s, directed along −x (that is, 0.314 m/s).
  • aₓ = −100π² sin π = 0
  • a_y = −100π² cos 2π = −986.96 mm/s²
  • a = 986.96 mm/s² ≈ 0.987 m/s², directed along −y.

(b) Path Since cos 2πt = 1 − 2 sin²πt, y = 25(1 − 2 sin²πt) = 25 − 50 (x/100)².

y = 25 − x²/200 (mm)

This is a parabola with its vertex at (0, 25) mm, opening downward. The particle oscillates along the arc between x = −100 mm and x = +100 mm (y from +25 mm down to −25 mm), repeating every 2 s. It passes the vertex (0, 25) at t = 0, 1 s and 2 s, and reaches the ends of the arc, (100, −25) and (−100, −25), at t = 0.5 s and 1.5 s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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