Amass 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 .

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

Round your answer to three decimal places."

The spring system is determined by the following differential equation

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

Solving for the damping coefficient.

where the spring constant k is given by

And the mass is given by

So the damping coefficient will be

Replacing the given information we have:

Thus we get

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

For this case we must convert 26 seconds to hours,

By definition we have to:

1 hour equals 3600 seconds.

So, if we propose a rule of three we have:

1 hour ----------> 3600 seconds

x ------------------> 26 seconds

Where the variable "x" represents the number of hours equivalent to 26 seconds:

Thus, 26 seconds equals approximately 0.0072 hours.

0.0072 hours