Answer:
a)
[tex]u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)[/tex]
b)
m = 48lb
c)
b = 144.76lb
Explanation:
The general equation of a damping oscillate motion is given by:
[tex]u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)[/tex] (1)
uo: initial position
m: mass of the block
b: damping coefficient
w: angular frequency
α: initial phase
a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:
[tex]|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s[/tex]
Then, you obtain by replacing in (1):
6in = 0.499 ft
[tex]u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)[/tex]
b.
mass, m = 48lb
c.
b = 144.76 lb/s
