A particle starts from the origin at t = 0 and moves along the positive x axis.

Question

A particle starts from the origin att= 0 and moves along the positivexaxis. A graph of the velocity of the particle as a function of the time is shown in the figure; thev-axis scale is set byvs= 5.0 m/s.(a)What is the coordinate of the particle att= 5.0 s?(b)What is the velocity of the particle att= 5.0 s?(c)What is the acceleration of the particle att= 5.0 s?(d)What is the average velocity of the particle betweent= 1.0 s andt= 5.0 s?(e)What is the average acceleration of the particle betweent= 1.0 s andt= 5.0 s?

Explanation / Answer

A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2.10 cm, and the frequency is 1.30 Hz. Potential energy = ½ * k * (maximum distance stretched)^2 Amplitude is the distance the particle moves in one direction. maximum distance stretched = amplitude = 2.10 cm = 0.021 m As the spring expands, the potential energy decreases and the kinetic energy increases. (b) Determine the maximum speed of the particle. Maximum Potential energy = ½ * k * (0.021)^2 As the spring expands, the potential energy decreases and the kinetic energy increases. Maximum Kinetic energy = ½ * mass * (maximum velocity)^2 Increase of Kinetic energy = decrease of Potential energy ½ * mass * (maximum velocity)^2 = ½ * k * (0.021)^2 (maximum velocity)^2 = (k/m) * (0.021)^2 Frequency is number times that the particle moves back and forth each second. Period is the time of one oscillation = 1/Frequency = (1/1.30) seconds The equation below shows the relationship between time, mass, and the elasticity coefficient. Time = 2 * p * (m/k)^0.5 (1/1.30) = 2 * p * (m/k)^0.5 multiply both sides by 1.30 1 = 1.30 * 2 * p * (m/k)^0.5 (m/k)^0.5 = 1/(1.30 * 2 * p) square both sides m/k = 1/(1.30 * 2 * p)^2 (maximum velocity)^2 = (k/m) * (0.021)^2 k/m = (1.30 * 2 * p)^2 (maximum velocity)^2 = (1.30 * 2 * p)^2 * (0.021)^2 maximum velocity = 1.30 * 2 * p * 0.021 (c) Determine the earliest time (t > 0) at which the particle has this speed The particle has maximum speed when the spring has expanded 0.021 meters = amplitude = ½ * Time for complete cycle Time for complete cycle = Period = 1/1.30 seconds Time = ½ * 1/1.30 = (1/2.6) seconds (d) Find the maximum positive acceleration of the particle. The maximum positive acceleration occurs when the force is maximum Average force * maximum distance stretched = PE max Maximum Potential energy = ½ * k * (0.021)^2 k/m = (1.30 * 2 * p)^2 k = m * (1.30 * 2 * p)^2 Average force * 0.021 = ½ * m * (1.30 * 2 * p)^2 * (0.021)^2 Maximum force = 2 * Average force Maximum force = m * (1.30 * 2 * p)^2 * (0.021)^2 maximum positive acceleration = Maximum force * mass maximum positive acceleration = (1.30 * 2 * p)^2 * (0.021)^2 (e) Find the earliest time (t > 0) at which the particle has this acceleration. The time at which particle has this acceleration = time for ½ of a cycle. Time = ½ * 1/1.30 = (1/2.6) seconds (f) Determine the total distance traveled between t = 0 and t = 1.15 s = Each (1/2.6) seconds, the particle moves the amplitude = 0.021 m 1.15 ÷ (1/2.6) = 2.99 amplitudes = 2.99 * 0.021 = 0.06279 m Distance = 6.279 cm

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