Answer:
[tex]x(t) = e^-2t(c_1cos4t+c_2cos4t)[/tex]
[tex]x(t) = \sqrt{5}/2* e^-2tsin(4t+4.24)[/tex]
t =1.294
Explanation:
Let m be the mass attached, let k be the spring constant and let p be the positive damping constant. The Newton's second law for the system is
[tex]md^2x/dt^2 = -kx-\beta dx/dt[/tex]
where x(t) is the displacement from the equilibrium position. The equation can be transformed into
[tex]d^2x/dt^2 +\beta /m*(dx/dt)+k/m*x=0[/tex]
(a) Let's determine the equation of motion. We must convert units of weight into units of mass
[tex]m=W/g = 3.2/32 =1/10 slug[/tex]
From Hooke's law we can calculate the spring constant k.
[tex]k = W/s = 2/1 = 2lb/ft[/tex]
If we put m = 1/10 slugs, k = 2 lb/ft and [tex]\beta[/tex] = 0.4 into the DE, we get
[tex]d^2x/dt^2 +4*(dx/dt)+20*x=0[/tex]
The auxiliary equation is m^2+4*m+20 = 0 and its solutions are m_1 = -2-4_i and m_2 = -2 + 4i. The general solution is then
[tex]x(t) = e^-2t(c_1cos4t+c_2cos4t)[/tex]
From the initial conditions
[tex]x(0) = -1ft , x'(0) = 0ft/s[/tex]
we can find the equation of motion
[tex]x(t) = e^-2t(-cos4t-1/2sin4t)[/tex]
(b) We need to express the equation of motion in the alternative form
first attachment
The amplitude is
[tex]A=\sqrt{c_1^2+c_2^2}=\sqrt{(-1)^2+(-1/2)^2}= \sqrt{5}/2[/tex]
and the phase angle is
sin∅=[tex]c_1/A=-2/\sqrt{5}[/tex]
cos∅=[tex]c_2/A=-1/\sqrt{5}[/tex]
tan∅=2
∅=[tex]\pi +tan^-1*2[/tex]
=4.24
[tex]x(t) = \sqrt{5}/2* e^-2tsin(4t+4.24)[/tex]
(c) lb find the wanted time, we will need the derivative of the equation of motion
x'(t)=[tex]5*e^-2tsin4t[/tex]
The mass passes through the equilibrium when
x(t) =0
[tex]e^-2t(-cos4t-1/2sin4t)=0\\-cos4t-1/2sin4t=0\\tan4t=-2\\4t=-tan^-1+k\pi[/tex]
t =-0.2786+k[tex]\pi[/tex]/4
The velocity at those times is
second attachment
The first (positive) time at which the mass passes through the equilibrium heading upward is for k = 2 (when x' < 0).
t =1.294
