A particle moving along the x axis in simple harmonic motion starts from its equ

Question

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 2.20 cm and the frequency is 1.40 Hz.

(a) Show that the position of the particle is given by x = (2.20)sin(2.80t). (Do this on paper. Your instructor may ask you to turn in this work.)

(b) Determine the maximum speed and the earliest time (t > 0) at which the particle has this speed. cm/s s

(c) Determine the maximum acceleration and the earliest time (t > 0) at which the particle has this acceleration. cm/s2 s

(d) Determine the total distance traveled between t = 0 and t = 1.07 s.

Explanation / Answer

(a) w = 2.20f = 2.80

x = A sin wt

x = 2.20sin 2.80t

b) Vmax = Aw = 0.0220 x 2.80x = 0.193 m/s

after T/2.20 = 1/2.80 rd of a second ( 0.785s )

c) amax = A w2 = 1.776 = 1.700 m/s2

after T/4 = 1/6 th of a second ( 0.666 s )

d) Distance traveled = 12 x 1.07=12.84 cm

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