Respuesta :
Answer:
y = - 0.050 sin (131.59t )
Explanation:
In this exercise we are told to approximate the movement of a piston to the simple harmonic movement
y = A cos (wt + Ф)
in this case they indicate that the stroke (C) of the piston is twice the amplitude
C = 2A
A = C / 2
angular velocity is related to frequency
w = 2π f
let's substitute
y = [tex]\frac{C}{2}[/tex] cos (2π f t +Ф)
To find the phase (fi) we will use the initial conditions, the piston starts at the midpoint of the stroke, if we create a reference system where the origin is at this point
y = 0 for t = 0
we substitute in the equation
0 = \frac{C}{2} cos (0 + Ф)
The we sew zero values for the angles of Ф = π/2 rad
we substitute in the initial equation
y = \frac{C}{2} cos (2π f t + π/2)
let's use the double angle relationship
cos ( a +90) = cos a cos 90 - sin a sin 90
cos (a+90) = - sin a
y = -\frac{C}{2} sin (2πf t )
let's reduce the frequency to SI units
f = 200 rpm (2π rad / 1rev) (1 min / 60s) = 20.94 rad / s
we substitute the given values
y = - [tex]\frac{0.100}{2}[/tex] sin (2π 20.94 t )
y = - 0.050 sin (131.59t )
