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Fall 2008 -- Purdue University -- West Lafayette, IN
Blog Post LABELS
- Blog Challenge Questions (9)
- Chapter 2 (2)
- Chapter 3 (19)
- Chapter 5 (11)
- Chapter 6 (23)
- Chapter 7 (7)
- Chapter 8 (9)
- Exam 1 (5)
- Exam 2 (4)
- Exam 3 (17)
- Exams (2)
- Final Exam (7)
- General (6)
- Gravity (2)
- Homework (2)
- in class (1)
- Lectures (1)
- Quizzes (2)
- Spring 2008 Archive (87)
- Spring-Mass Simulator (1)
- Student-Generated Solutions (1)
- Summer 2008 Archive (72)
12 comments:
when you set up your forces, you know that the distance traveled by the mass, y, is half that traveled by B, yb. I hope this helps.
Why is the distance traveled by the mass half of that traveled by B?
because y travels twice as much as yb your spring force is k(2y-yb). Also don't forget because it's a pulley you have the same force acting on the other end so have this force twice. Therefore your EOM is -2k(2y-yb)=m*y''.
Do we even have to find the particular solution to find omega c or do we merely find it using the same method we have been using to find omega n?
you will use the same method as to find omega_c as we have for omega_n
^^ Is this true for only undamped, forced vibration cases?
In regards to M Sandercock's post, we did that in class Monday in a problem in which "the ocsillations of the mass became excessively large." Does that have anything to do with Wc = Wn, or is it solely that it is undamped?
kyle/janet: thanks, that helped.
M Sandercock: Since the question is asking for when the oscillations are excessively large, this will always happen when
omega_c = omega_n
so i think even if there is a dashpot, it would still be the square root of whatever is in front of the x (or y) term.
I know this was discussed in lecture, but I'm still confused on the explanation. For this type of problem, when is it necessary to include the force of gravity in the EOM, and when can it be neglected?
When the displacement, x, is measured from the position where the spring is unstretched, you will naturally get the weight term in the EOM.
When the displacement, z, is measured from the position where the mass is in static equilibrium, the weight does NOT appear in the EOM.
Why is this? One explanation that we talked about in class is that z and x are related by z = x - x_static, where x_static is the value of x at static equilibrium, or x_static = m*g/k. If you substitute this into the EOM in terms of x, you will get the EOM in terms of z with the weight term canceling out. A more physical explanation is that weight influences the position of mass at static equilibrium (where the spring forces balance out the weight forces). Weight does NOT, however, influence the dynamic (or vibrational) component of the response.
So, in summary, weight appears when you measure position from the unstretched state, and it does NOT appear when you measure position from equilibrium. In this problem, the figure shows that y is measured from static equilibrium, and therefore weight does not appear.
I hope that this helps. If not, let us know.
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