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theBEAST
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Homework Statement
My team and I are working on a design project to design/modify a device that can go on hikes for paraplegic/quadriplegic people.
Here is the current design (not designed by us):
http://i.imgur.com/RAOq28k.png
http://i.imgur.com/zgA3aCg.png
We are thinking of adding a spring damping system to reduce shock on the passenger. We calculated that we would need a k = 87.3 kN/m and b = 4.137 kNs/m to meet some of the specifications. The mass of the entire device is about 100 kg. Using these we want to find the acceleration felt by the rider after going over a bump. So we decided to model it using an impulse. We calculated that on average the bumps would create an acceleration of 1.36 m/s^2 on the rider without a damping system. Thus, we want to find out what the actual acceleration would be with a spring and damper.
The Attempt at a Solution
Plugging this into wolfram gives us:
x(t) = .064461*e^(-20.685t)*sin21.098t
(http://www.wolframalpha.com/input/?i=x''+++41.37*x'+++873*x+=+1.36*delta(t),+x(0)+=+0,+x'(0)+=+0)
Taking the second derivative gives us the acceleration:
x''(t) e^(-20.685 t) (-56.2631 cos(21.098 t)-1.11236 sin(21.098 t))
(http://www.wolframalpha.com/input/?i=second+derivative+of+.064461*e^(-20.685t)*sin21.098t)
This means that the magnitude of the maximum acceleration will be 56.3 m/s^2 which does not make sense since the input was only 1.36 m/s^2. Any idea what is going on here?
In fact I found the solution to x(t) online and it corresponds with what I got from wolfram: