Respuesta :
Answer:
The motion is over-damped when λ^2 - w^2 > 0 or when [tex]b^{2}[/tex] > 0.86
The motion is critically when λ^2 - w^2 = 0 or when [tex]b^{2}[/tex] = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when [tex]b^{2}[/tex] < 0.86
Explanation:
Using the newton second law
k is the spring constante
b positive damping constant
m mass attached
[tex]m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}[/tex]
x(t) is the displacement from the equilibrium position
[tex]\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0[/tex]
Converting units of weights in units of mass (equation of motion)
[tex]m = \frac{W}{g} = \frac{14}{32} = 0.43 slug[/tex]
From hook's law we can calculate the spring constant k
[tex]k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft[/tex]
If we put m and k into the DE, we get
[tex]\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0[/tex]
Denoting the constants
2λ = [tex]\frac{b}{m}[/tex] = [tex]\frac{b}{0.43}[/tex]
λ = b/0.215
[tex]w^{2} = \frac{k}{m} = 16.28[/tex]
λ^2 - w^2 = [tex]\frac{b^{2} }{0.046} - 16.28[/tex]
This way,
The motion is over-damped when λ^2 - w^2 > 0 or when [tex]b^{2}[/tex] > 0.86
The motion is critically when λ^2 - w^2 = 0 or when [tex]b^{2}[/tex] = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when [tex]b^{2}[/tex] < 0.86
(1) The motion is over-damped when [tex]\lambda^2-w^2 > 0[/tex]λ^2 - w^2 > 0 or when [tex]b^2[/tex]> 0.86
(2) The motion is critically when [tex]\lambda^2-w^2 =0[/tex] or when [tex]b^2[/tex] = 0.86
(3) The motion is under-damped when [tex]\lambda^2-w^2 > 0[/tex] or when [tex]b^2[/tex]< 0.86
What will be the motion of the spring-mass system?
From newton's second law of motion
k = Spring constant
b = positive damping constant
m =mass attached
[tex]m\dfrac{d^2x}{dt^2} =-kx-b\dfrac{dx}{dt}[/tex]
Since x(t) is the displacement from the equilibrium position
[tex]\dfrac{d^2x}{dt^2} +\dfrac{h}{m} \dfrac{dx}{dt} +\dfrac{k}{m} x=0[/tex]
Converting units of weights in units of mass (equation of motion)
[tex]m=\dfrac{w}{g} =\dfrac{14}{32} =0.43[/tex]
From hook's law, we can calculate the spring constant k
[tex]k=\dfrac{w}{s} =\dfrac{14}{2} =7 \frac{lb}{ft^2}[/tex]
If we put m and k into the equation we get
[tex]\dfrac{d^2x}{dt^2} +\dfrac{b}{0.43} \dfrac{dx}{dt} +16.28x=0[/tex]
Denoting the constants
[tex]2\lambda=\dfrac{b}{m} =\dfrac{b}{0.43}[/tex]
[tex]\lambda =\dfrac{b}{0.215}[/tex]
[tex]\lambda^2-w^2=\dfrac{b^2}{0.046} =16.28[/tex]
Thus
(1) The motion is over-damped when [tex]\lambda^2-w^2 > 0[/tex]λ^2 - w^2 > 0 or when [tex]b^2[/tex]> 0.86
(2) The motion is critically when [tex]\lambda^2-w^2 =0[/tex] or when [tex]b^2[/tex] = 0.86
(3) The motion is under-damped when [tex]\lambda^2-w^2 > 0[/tex] or when [tex]b^2[/tex]< 0.86
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