A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point foot above the equilibrium position with a downward velocity of ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c)  3e^-4/3

Explanation:

Given that

g = 32 ft/s^2 ,  spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity  hence  β = 1

a)Calculate the time at which the mass passes through the equilibrium position.

time mass passes through equilibrium = 1/12 seconds = 0.083

b) Calculate the time at which the mass attains its extreme displacement

time when mass attains extreme displacement = 1/3 seconds = 0.33 secs

c) What is the position of the mass at this instant

position = 3e^-4/3

attached below is the detailed solution to the given problem

he . if i may ask what i your question exactly?

explanation:

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.