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.)

The motion is over-damped when λ^2 - w^2 > 0 or when > 0.86

The motion is critically when λ^2 - w^2 = 0 or when = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

x(t) is the displacement from the equilibrium position

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

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

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

Denoting the constants

2λ = =

λ = b/0.215

λ^2 - w^2 =

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when > 0.86

The motion is critically when λ^2 - w^2 = 0 or when = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

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