Welcome to the website of ME 274 for the Fall 2008 semester. On this site you can view blog posts, add your own blog posts and add comments to existing posts. In addition to the blog are links to course material: course information, information on solution videos, exams, quizzes, homeworks and other course-related material. Direct links to the homework solution videos are also available on the left side of this page.


The following is a reverse chronological order listing of the posts for the course blog. To add a post, click here (when adding posts, be sure to add a "label" in the box at the lower right side of the post window). To add a comment to an existing post, click on the "Comments" link below the post.


____________________________________________________

Apr 9, 2008

Problem 8/9

In this problem, I have the frequency equal to sqrt(k/m). If this is correct how do I go about solving for k so that i can solve for the frequency itself? Thanks.
Posted by at
Labels:

4 comments:

CMK said...

Note that the problem statement gives us the static deformation, x_st, of the spring under the weight of the block. In lecture, we saw that: x_st = mg/k. You can use this equation to find k. From that, you can find the natural frequency.

[Yes, in THIS case, omega_n = sqrt(k/m), where omega_n is in rad/sec. Also, f_n (in Hz) = omega_n/(2*pi).]

April 9, 2008 at 7:47 PM
lkirtiad said...

I got omega_n=sqrt(grav/x_st) which yields to omega_n equals 19.8 (from square root of 9.81/25E-3) . Is this correct?

I found that this was a free vibration problem, so
x(t)=Acos(omega_n*t)+Bsin(omega_n*t)
how do we find B?

April 10, 2008 at 2:33 AM
Chris Cohoat said...

Your omega_n has to be computed with static deflection in meters. In the final equation, x0 is in millimeters.

Also, B = x0_dot/wn. As I understood it, x0_dot was zero in this instance because it was measured from the instant of release.

April 11, 2008 at 10:19 AM
Chris Cohoat said...

Also, .05 should be your static deflection. It looks like you used 25mm (the added deflection) instead of the static deflection.

April 11, 2008 at 10:22 AM

Post a Comment

Subscribe to: Post Comments (Atom)