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 motions is 2.00 cm, and the frequency is 1.50 Hz
a) Show that the position of the particle is given by x=(2.00 cm)sin(3.00?t)
b) Determine the maximum speed and the earliest time (t >0) at which the particle has this speed
c) The maximum acceleration and the earliest time (t>0) at which the particle has this acceleration
d) The total distance traveled between t=0 and t=1.00 s

Explanation / Answer

its motion should be in the form : x(t) = 2.1 sin (2pi f) t => x(t) = 2.1 sin 8.17 t so its speed is : v(t) = dx/dt = 8.17(2.1) cos 8.17t => v(t) = 17.16 cos 8.17t a) Vmax = 17.16 m/s b) here u should have 8.17 t = pi = 3.14 approx so that |cos 8.17 t| = 1 => t = 0.3845 s c) a(t) = dv/dt = - 17.16 (8.17) sin 8.17t = -140.2 sin 8.17t d) a(t) max = 140.2 m/s^2 e) 8.17t must equal (3pi/2) in order to get a = + 140.2 => 8.17 t = 3pi/2 = 4.7 => t = 0.576 s f) 2.1 sin [8.17 (1.15)] - 0 = 0.0615 m

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