Question

A mass weighing 30 lbstretches a spring . The mass is also attached to a damper with coefficient . Determine the value of for which the system is critically damped. Assume that .

Answer 1

Answer:

the damping coefficient when the system is critically damped is 13.42 lb s/ft.

Explanation:

The directions are missing the last part, which is:

"A mass weighing 30 lb stretches a spring 7.5 in. The mass is also attached to a damper with coefficient γ. Determine the value of γ for which the system is critically damped. Assume that $g = 32 \cfrac {ft}{s^2}$

Round your answer to three decimal places."

The spring system is determined by the following differential equation

$my''+\gamma y'+ky=0$

where $\gamma$ is the damping coefficient, thus it is critically damped when the system transitions from real to complex solutions that happens at

$\gamma_^2 -4km=0$

Solving for the damping coefficient.

$\gamma^2 = 4km\\ \gamma = \sqrt{4km}$

where the spring constant k is given by

$k = \cfrac{mg}{L}\\k = \cfrac wL$

And the mass is given by

$mg = w\\ m = \cfrac wg$

So the damping coefficient will be

$\gamma = \sqrt{4\cfrac wL \cfrac wg}\\\gamma = \sqrt{4\cfrac {w^2}{gL}}$

Replacing the given information we have:

$\gamma = \sqrt{4\cfrac {(30\, lbs)^2}{32\cfrac{ft}{s^2}0.625 ft}$

Thus we get

$\gamma = 13.42 \cfrac{lb \cdot s}{ft}$

The value of the damping coefficient when the system is critically damped is 13.42 lb s/ft.