Homework 6, due October 15, 2014

Due at the beginning of class, Wednesday, October 15, 2014. No extension due to exam on Thursday, October 16, 2014.
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goodwine
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Homework 6, due October 15, 2014

Post by goodwine »

Reading: By now you should have read all of chapters 1 - 4 except section 1.6 from Chapter 1.

Exercises: 4.17, 4.19, 4.26 and 4.27.
Bill Goodwine, 376 Fitzpatrick
smohdyaz

Re: Homework 6, due October 15, 2014

Post by smohdyaz »

For question 4.26, is the F in the diagram equal to mg, the 3EIx/L^3, or both?
goodwine
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Re: Homework 6, due October 15, 2014

Post by goodwine »

smohdyaz wrote:For question 4.26, is the F in the diagram equal to mg, the 3EIx/L^3, or both?
Neither. If you want to deflect the beam by an amount x it takes a force of magnitude 3IEx/L^3. The problem is asking of a mass is on the end how does it vibrate? You may ignore the mass of the beam itself and, thanks to 4.9 from last week, you may ignore gravity because we now know that if we set x=0 at equilibrium, gravity does not appear in the equation of motion.
Bill Goodwine, 376 Fitzpatrick
lpaquin

Re: Homework 6, due October 15, 2014

Post by lpaquin »

For 4.27, I am getting that f(t) = mrw^2*sin(wt).
When m is aligned on the x axis, theta=0, and the centripetal force is entirely horizontal so I would assume that you would multiply mrw^2 by cos(wt) for the horizontal component and sin(wt) for the vertical component. Am I measuring theta from the wrong axis?
goodwine
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Re: Homework 6, due October 15, 2014

Post by goodwine »

lpaquin wrote:For 4.27, I am getting that f(t) = mrw^2*sin(wt).
When m is aligned on the x axis, theta=0, and the centripetal force is entirely horizontal so I would assume that you would multiply mrw^2 by cos(wt) for the horizontal component and sin(wt) for the vertical component. Am I measuring theta from the wrong axis?
The figure doesn't really say where t=0, so it can be either sine or cosine.
Bill Goodwine, 376 Fitzpatrick
mhughan

Re: Homework 6, due October 15, 2014

Post by mhughan »

For 4.27, part 2, can we assume that "steady state" indicates that there is only a particular solution even though our damping ratio is zero?
goodwine
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Re: Homework 6, due October 15, 2014

Post by goodwine »

mhughan wrote:For 4.27, part 2, can we assume that "steady state" indicates that there is only a particular solution even though our damping ratio is zero?
Good question. The answer is yes.
Bill Goodwine, 376 Fitzpatrick
smohdyaz

Re: Homework 6, due October 15, 2014

Post by smohdyaz »

goodwine wrote:
mhughan wrote:For 4.27, part 2, can we assume that "steady state" indicates that there is only a particular solution even though our damping ratio is zero?
Good question. The answer is yes.
Why is that still valid even without the exponential term of the x_h?
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Re: Homework 6, due October 15, 2014

Post by goodwine »

smohdyaz wrote:
goodwine wrote:
mhughan wrote:For 4.27, part 2, can we assume that "steady state" indicates that there is only a particular solution even though our damping ratio is zero?
Good question. The answer is yes.
Why is that still valid even without the exponential term of the x_h?
Good question. The mathematical answer is that the homogeneous solution is still there forever without damping. However, in the case where the particular solution is very large (near resonance), then it would be justifiable to ignore it. It's not uncommon in vibrations to ignore damping because doing so is sort of conservative, and still just study the particular solution. Doing so is justified if there is even a small amount of damping and/or if the particular solution is large compared to the homogeneous solution. Also, we never really know the initial conditions anyway, so focusing on the particular solution in that case makes the most sense as well.
Bill Goodwine, 376 Fitzpatrick
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