A mass weighing 13 lb stretches a spring 1.5 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 ft/s2. Round your answer to three decimal places.

γ = 13 lb.s/ft

Step-by-step explanation:

Weight = 13 lb

L = 1.5 inches = 0.125 ft

To determine the value of γ for which the system is critically damped.

We write the general equation for such a system

my" + γy' + ky = 0

And this is critically damped when the roots are equal, that is

γ² - 4km = 0

k = spring constant = (weight/L) = (13/0.125) = 104 lb/ft

m = (w/g) = (13/32.2) = 0.404 lb.s²/ft

γ² - 4km = 0

γ² - 4(104)(0.404) = 0

γ² = 168.604

γ = 12.98 = 13 lb.s/ft

21.4 i did it already for s1

step-by-step explanation:

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )