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 1.60 cm, and the frequency is 2.70 Hz.

(a) Find an expression for the position of the particle as a function of time. (Use the following as necessary: t, and ?.)
x =  

(b) Determine the maximum speed of the particle.
cm/s

(c) Determine the earliest time (t > 0) at which the particle has this speed
s

(d) Find the maximum positive acceleration of the particle.
cm/s2

(e) Find the earliest time (t > 0) at which the particle has this acceleration.
s

(f) Determine the total distance traveled between t = 0 and t = 0.56 s.

Explanation / Answer

ts motion should be in the form :

x(t) = 1.6 sin (2pi f) t

=> x(t) = 1.6 sin 16.96 t

so its speed is :

v(t) = dx/dt = 16.96(1.6) cos 16.96 t

=> v(t) = 27.14 cos 16.96 t

a) Vmax = 27.14 m/s

b) here u should have 16.96 t = pi = 3.14 approx so that |cos 16.96 t| = 1

=> t = 0.185 s

c) a(t) = dv/dt = - 27.14 (16.96) sin 16.96t = -460.29 sin 16.96 t

d) a(t) max = 460.96 m/s^2

e) 16.96 t must equal (3pi/2) in order to get a = + 140.2

=> 16.96 t = 3pi/2 = 4.7
=> t = 0.277 s

f) 1.6 sin [ 16.96*(0.56)] - 0 = 0.264 m

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