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Tuesday, December 4, 2007

problem 8-61

[Sketch added by CMK.]


i have a question about the answers. While i see how to get them, i'm not entirely sure what the answers imply, and how if we have w/wn > SQRT(2) it would have an amplitude less than 2b. Can anyone help me understand the physical aspect of the problem?

8 comments:

CMK said...

If you plot Z(omega) (the amplitude of forced response for the relative coordinate z(t) defined in lecture) you will see that the response is largest near omega = omega_n since that corresponds to near-resonant response. (See a sketch of Z(omega) that I have provided in your original post.)

If you move to frequencies of excitation that are either higher or lower than omega_n, the amplitude of response for z(t) drops off since you are moving away from resonance. The two excitation frequencies that you are asked to find (omega_1 and omega_2 shown in the sketch) are ones for which the amplitude of response (in absolute value) is less that 2b to the left of omega_1 and to the right of omega_2.

Does this help?

lemerson said...

Could you help me get as far as getting the solutions in the book. I can't get an equation that relates them without involving sin.

CMK said...

* Find the particular solution to the EOM that we derived in lecture: x(t) = X*sin(omega*t)
* Find the response relative to the cart, as defined in lecture: z(t) = x(t) - xB(t) = Z*sin(omega*t)
* The plot of Z(omega) is shown in the figure attached to John's original post.
* Find the two frequencies omega_1 and omega_2 shown in the sketch.

Wendy said...

from the EOM we got in class,
x_dot_dot + k/m x = kb/m sin(wt), which is in form of equation 8/15, with F_o = kb. and then I used equation 8/17 to find the amplitude, X = (F_o/k)/[1-(w/w_n)^2], and got w/w_n < sqrt(1/2).
why is this wrong?

CMK said...

Note that you are asked to find the excitation frequencies for which the amplitude (absolute value) of the motion of the block RELATIVE to the cart is less than 2b.

To find the relative motion, use the coordinate z(t) = x(t) - xB(t) = Z*sin(omega*t) as discussed in lecture.

A sketch of Z(omega) vs. omega is shown in John's original post. From this plot, we see that there are two frequency ranges for which the absolute value of Z is less than 2b. See above comment.

Let me know if this does not help.

Deborah said...

what do we do with the picture I am not understanding how to find the ratios?

CMK said...

The figure is intended to show that the magnitude of the amplitude of relative motion, Z, is less than 2b for omega < omega_1 and for omega > omega_2.

From this you can see that omega_1 corresponds to Z = 2b and that omega_2 corresponds to Z = -2b.

Anonymous said...

I feel pretty helpless here. Are we to use the amplitude equation( X = (F_o/k)/[1-(w/w_n)^2]) at all. Are we to substitute those eqns? I'm really lost.