The mass of the system shown is released from rest at x 0 = 5.9 in. when t = 0.
ID: 1828353 • Letter: T
Question
The mass of the system shown is released from rest at x0 = 5.9 in. when t = 0. Determine the displacement x att = 0.46 sec if (a) c = 8.8 lb-sec/ft and (b) c = 11.9 lb-sec/ft
Here is a link to the diagram/problem:
http://i1334.photobucket.com/albums/w655/salonemark/8-38_zps1eac6537.jpg
The answer must be in Inches and please calculate out the number.
Answer is asked as follows:
(a) If c = 8.8 lb-sec/ft and t = 0.46 sec: x = ____ inches
(b) If c = 11.9 lb-sec/ft and t = 0.46 sec: x= ____inches
The mass of the system shown is released from rest at x0 = 5.9 in. when t = 0. Determine the displacement x att = 0.46 sec if (a) c = 8.8 lb-sec/ft and (b) c = 11.9 lb-sec/ft The answer must be in Inches and please calculate out the number. If c = 8.8 lb-sec/ft and t = 0.46 sec: x = inches If c = 11.9 lb-sec/ft and t = 0.46 sec: x= inchesExplanation / Answer
the equation for this arrangement will be
mx'' - cx' - kx = 0
x = displacement
x' = velocity = dx/dt
x'' = acceleration = d2x/dt2
64.4x'' - 8.8x' - 9.68x = 0
interating once wrt time t
64.4x' - 8.8x - 9.68x^2/2 = C
at t = 0 ,velocity is zero
x' = 0
putting x = 5.9 in = .492 ft
C = -5.5
integrating again
64.4x - 8.8x^2/2 - 4.84x^3/3 = Ct + D ...................................2
now as per the given condition
at t = 0 , x = 5.9 in = 0.492 ft
putting this in 2 we get
D = 30.43
so the equation becomes
64.4x - 8.8x^2/2 - 4.84x^3/3 = -5.5t + 30.43
now putting t = 0.46 s
64.4x - 4.4x^2 - 1.61x^3 -27.9 = 0
solving for x
x = 0.4492 ft
x = 5.3916 in
in case the c = 11.9
equations
64.4x' - 11.9x - 4.84x^2 = C
64.4x - 5.95x^2 - 1.61x^3 = Ct + D
puttin t = 0 ,x = .492 ft ,x' = 0
C = -7.03
D = 30.05
equation becomes
64.4x - 5.95x^2 - 1.61x^3 +7.03t - 30.05 = 0
putting t = 0.46
solving for x
x = 0.4360 ft
x = 5.232 in
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