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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)
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- in class (1)
- Lectures (1)
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- Spring 2008 Archive (87)
- Spring-Mass Simulator (1)
- Student-Generated Solutions (1)
- Summer 2008 Archive (72)
13 comments:
I also had the same thoughts when approaching this problem. When subsituting the x_p'' and x_p into my EOM, I also had the gravitational term. I believe the solution to this problem is group the all 'like' terms. I then had the following:
sinwt: -mX_2*w^2 + kX_2 = 0
coswt: -mX_1*w^2 + k_X_1 = f_o
gravitational: -mg = 0
I hope this is helpful, the posted solution for problem 8/50 is also very helpful, but does not discuss the gravitational value.
I like Dale's approach but with one change. Write xp(t) = X1*cos(omega*t) + X2*sin(omega*t) + A. Substituting this into the EOM and grouping together like terms gives:
sine: -m*X2*omega^2 + k*X2 = 0
cosine: -m*X1*omega^2 + k*X1 = f0
constant: k*A = -m*g
This is the same as Dale's except the last equation. This last equation gives: A = -m*g/k.
An alternate approach would be to redefine the coordinate x as being measured from static equilibrium. As we have seen in class, and as is discussed in the text, the EOM in terms of such a coordinate does not contain the weight term.
The amplitude of oscillation for either approach is the same. With the first approach, the constant part of the response "A" does not affect the oscillations.
In the book's example, Sample Problem 8/6 (pg 628), they only consider the "dynamic forces" in their FBD and hence they disregard the mg term completely.
The rationale for including only dynamic forces is that only dynamic forces appear in the EOM when measuring x from the static equilibrium state.
I'm a little confused on this problem and feel that I'm probably overlooking something. I have X1 in terms of omega, omega_n, f_o and k. I'm not sure what value to use for omega; Or do we leave the final solution in terms of a variable, omega?
Phil, we can find a number solution for the problem using our equation for X, X = sqrt((x1)^2 + (x2)^2). From the above equations posted by CMK we see that x2 = 0. Therefore we have X = x1.
For the x1 equation we have the terms w, w_n, f_0, and k.
You can solve for the value of w_n using information from your EOM.
The value of w is given to us in the equation for f.
The value for f_0 can be found by using t=0 and k is a given value.
Plug all of these values into your equation for x1 and you have the answer.
Phil, omega is the coefficient of t in F in this case it's 20 since F is given as 1000*cos(20t)
I found that assuming the mass starts at static equilibrium makes the problem easier. Then you don't have to worry about the weight force.
Now for the end solution, I don't think we need omega_n. In the example class on page 27 the problem used omega in Xb.
Brian, I think you meant the coefficient of omega is 120 since the equation is F = 1000cos(120*t) N, rather than 20.
mtriebe,
you can get an omega_n^2 in the problem on page 27 if move the mass around. I think ending up with an omega_n in your end equation just depends on if you got rid of the coefficients of x_dot_dot in your EOM originally.
I get a negative vaue on the left side of the cosine equation, is this just due to the fact that the mass had moved lower than the equillibrium point. I assume the amplitude is just the absolute value of this number.
I see what you're saying about not needing w_n, but is that possible? I was looking at the example on page 628 in the text and it seems like we needed it though.
About omega_n, is it needed in the solution if you can just solve it with omega? I noticed in 8/50 that the professor divided by k to get omega_n in the answer but before that everything was known. Is there a need to divide by k to get omega_n in the answer?
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