---
title: "Vibrating particle velocity, acceleration and path"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:vibrating-particle-x-100-sin-pi-t"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/vibrating-particle-x-100-sin-pi-t
site: BCT Engineering
---

# Vibrating particle velocity, acceleration and path

Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body

- Asked in: 2070 Asar, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/vibrating-particle-x-100-sin-pi-t
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

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.

## 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).

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are v_x and v_y?
   A. v_x = 100π cos πt, v_y = -50π sin 2πt
   B. v_x = 100 cos πt, v_y = -25 sin 2πt
   C. v_x = -100π cos πt, v_y = 50π sin 2πt
   D. v_x = 100π sin πt, v_y = 25 cos 2πt
   Answer: A. Differentiate each equation using the chain rule. (Key point: Velocity components)

2. At t = 1 s, what is the velocity?
   A. 314.16 mm/s along +x
   B. 986.96 mm/s along -y
   C. Zero
   D. 314.16 mm/s along -x
   Answer: D. v_x = 100π cos π = -314.16 and v_y = -50π sin 2π = 0. (Key point: Velocity at t = 1)

3. What is the acceleration at t = 1 s?
   A. 986.96 mm/s^2 along +y
   B. 986.96 mm/s^2 along -y
   C. 314.16 mm/s^2
   D. Zero
   Answer: B. a_x = -100π^2 sin π = 0 and a_y = -100π^2 cos 2π = -986.96. (Key point: Acceleration at t = 1)

4. What is the path and how is it found?
   A. Circle, since sin^2 + cos^2 = 1
   B. Ellipse
   C. Parabola: using cos 2πt = 1 - 2 sin^2 πt gives y = 25 - x^2/200
   D. Straight line
   Answer: C. Eliminating t gives y = 25 - x^2/200. (Key point: Path)

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