Question

A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the mass has a velocity of 6 ft/sec. Suppose the object is displaced an additional 6 inches and released.
Required:
a. Find an equation for the object's displacement, u(t), in feet after t seconds.
b. What is the mass of the object?
c. What is the damping coefficient?
d. What is the spring constant?

Answer 1

Answer:

a)

$u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)$

b)

m = 48lb

c)

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

$u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)$    (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:

$|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s$

Then, you obtain by replacing in (1):

6in = 0.499 ft

$u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)$

b.

mass, m = 48lb

c.

b = 144.76 lb/s