Question

A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numerically equal to β (β > 0) times the instantaneous velocity. Determine the values of the damping constant β so that the subsequent motion is overdamped, critically damped, and underdamped. (If an answer is an interval, use interval notation. Use g = 32 ft/s2 for the acceleration due to gravity.)

Answer 1

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when $b^{2}$ > 0.86

The motion is critically when λ^2 - w^2 = 0 or when $b^{2}$ = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when $b^{2}$ < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

$m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}$

x(t) is the displacement from the equilibrium position

$\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0$

Converting units of weights in units of mass (equation of motion)

$m = \frac{W}{g} = \frac{14}{32} = 0.43 slug$

From hook's law we can calculate the spring constant k

$k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft$

If we put m and k into the DE, we get

$\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0$

Denoting the constants

2λ = $\frac{b}{m}$ = $\frac{b}{0.43}$

λ = b/0.215

$w^{2} = \frac{k}{m} = 16.28$

λ^2 - w^2 = $\frac{b^{2} }{0.046} - 16.28$

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when $b^{2}$ > 0.86

The motion is critically when λ^2 - w^2 = 0 or when $b^{2}$ = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when $b^{2}$ < 0.86