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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)
11 comments:
This amplitude, A, must be between -2b and 2b, so you can set up an inequality with this information and solve for omega in terms of k and m. I was also slightly confused as to why the book decided to introduce omega_n, but you can easily get the shown answer by dividing your inequality by omega_n which we found to be sqrt(k/m). Perhaps the book did this merely to eliminate the k and m terms of the inequality.
On a related note, sort of - I almost got the correct answer according to the book, but I got sqrt(3/2) and 1/sqrt(2) instead of the inverse. Any ideas why?
Like Alex: I got the inverse of the answers; sqrt(1/2) and sqrt(3/2) (which is wrong), but what I did was..
you know: x(p) = A sin(wt)
you've solved for A. I turned my A into an expression like on page 622 (8/19). Then since you want to talk about maximums you can forget about oscillation. I then set my version of (8/19) equal to +2b and -2b. Then solved for w/w_n.
I hope this is close to right procedure, but I've made a mistake in there somewhere.. anyone for some blog pts care to point it out?
CMK for steal? :)
I had the same problem as Alex and J Biberstine. I came up with the inverse of the answers in the book and I can't figure out what would cause this problem.
I came up with
b/(1-(omega/omega_n)^2) < +-2b
the b's cancel, but i get 1/sqrt(2) for both positive and negative 2b... I think I'm even more lost than you guys. Help?
I'm having the same problem...is it possible that the book has a typo?
Adam - when the b's cancel, you'll end up with 1 on the left side. When you are doing the "-2b" part, make sure when you take it over to the left side that you add it and not subtract it.
anyone having any luck with this one yet? I am having the same problems.
Let me jump in on your discussion.
If you read the fine print of the question it asks about the "motion of the mass m RELATIVE to that of the cart". What does that mean?
It looks like all of you are on the right path for finding the motion of the mass: x(t) = A*sin(omega*t). From this, you need to next find the motion of the mass RELATIVE to the base: x(t) - xB(t) = A*sin(omega*t) - b*sin(omega*t) = (A-b)*sin(omega*t).
Next, you need to make sure that abs(A - b) < 2b (where "abs" means absolute value).
Does this help? If not, let us know.
I'm still a bit confused I thought when we solved for the particular solution we found B to be zero, is this a different B that we are using to find the relative omega?
I wasn't sure what to do about the abs(A-b) either, so I followed what the book did on page 622 (eq 8/17). If you find your EOM, you should be able to plug in b*(w/wn^2) for your Fo/k. Use this new value of b(w/wn)^2 / [1-(w/wn)^2] and set that less than your 2b. Using this inequality, I was able to get the correct answers.
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