A 4 LBfweight is attached to a spring suspended from the ceiling. When the weight comes to rest at equilibrium, the spring has been stretched 3 inches. The damping constant for the system is 2 LBf−sec ft . If the weight is raised 9 inches above equilibrium and given an initial upward velocity of 2 ft/sec, determine the equation of motion of the weight and give its damping factor, quasiperiod and quasifrequency.

attached below

Explanation:

The initial conditions :

x(0) = - 9 inches = -3/4 ft

x'(0) = - 2 ft/sec

spring stretched  3 inches = 1/4 ft

mass = w / g = 4 Ib / 32 ft/sec^2 = 1/8  slug

the spring constant ( k ) = w / l  = 4 / ( 1/4 ) = 16 Ib/ft

applying the second law of motion

m d^2x/dt^2 +  b dx/dt  + kx

= 1/8 d^2x/dt^2  + 2 dx/dt  + 16 x

= d^2x/dt^2  + 16 dx/dt  +  128 x  ------- ( 1 )

we will resolve the above equation to obtain the required equation of motion  x( t )

Attached below is the remaining part of the solution

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~