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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.

Sagot :

batolisis

Answer:

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

View image batolisis
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