Hello,

Currently you are planning to take Classical Dynamics, PHY 321. I would suggest that you look around for an inexpensive version of the text that we will be using. Also, Dr. Alexanian will be teaching the second half of the course using the same book. We will be using the 5^th edition of Classical Dynamics of Systems and Particles by Thornton and Marion. Below are some sites that have copies are less expensive prices. The text site is at http://www.thomsonedu.com/thomsonedu/student.do?topicid=71DA&sortby=copy&type=all_radio&courseid=PC09&product_isbn=0534408966&disciplinenumber=13

We will be starting with Chapter 2. There will be more on the course later in the summer. At some point I’ll be putting additional information at http://people.uncw.edu/hermanr/dynamics/

 I hope you are enjoying your summer. Try not to get too rusty!

Dr. Herman

Textbook sites – cheap versions

• http://www.bookbyte.com/product.aspx?isbn=0534408966&utm_source=googlebase&utm_medium=search

• http://www.amazon.com/gp/offer-listing/0534408966/ref=dp_olp_2/002-4628266-9143205

• http://product.half.ebay.com/Classical-Dynamics-of-Particles-and-Systems_W0QQprZ2386910QQtgZinfo

Solutions Manual

• http://product.half.ebay.com/Classic-Dynamics-of-Particles-and-Systems_W0QQprZ30510104QQtgZinfo

• http://groups.google.com/group/alt.math.undergrad/msg/8c0b42dc601719b1?

Here are some comments on the homework. I am sure you have played with the problems already. (You cannot wait until the last minute to start the homework and reading.)

#9 This problem is a little more involved than I was expecting. So, you might want to look at it last. If there is no external force, you have m x'' = -k(x-x0). Since x0 is constant, you can rewrite it as m (x-x0)'' = -k(x-x0). Ot , let y=x=x0 if you prefer: Then y''+omega^2 y=0. You can solve this.

Now, if there is a constant force, you have the new problem y''+omega y = F/m =const. To get the particular solution, you would not guess a cosine (why?). Try y=const = A. Insert into equation and find A. The general solution can be found. 

Now, the two problems apply for t=<t0 and t=>t0. You should have two different arbitrary constants for each solution that you have to find. The first set (last problem) you get from initial conditions. The second set you get by expecting that at t=t0 the two solutions and their first derivatives have to agree. This leads to the solution in the book. 

#28 I basically did this in class.  

#29 This is deceptive. The period is P = 4 Pi/omega. So, use the more general form for Fourier series: 

        F(t) = a0/2 + sum[an cos(2n pi t/P)+bn sin(2n pi t/P)] with  

       an = 2/P int_{-P/2}^{2P/2} F(t) cos(2n pi t/P) dt and bn = 2/P int_{-P/2}^{2P/2} F(t) sin(2n pi t/P) dt 

Plug in the above P and the given F. Note, you need to recall from Calc 2 how to integrate sin(ax)sin(bx) and sin(ax)cos(bx) using the product identities for trig functions.  

#40 Here you have a straightforward forced oscillation problem. mx''+kx = A sin omega t. Use the info to get the constants and use the known solution form to study the problem. 

#45 Recall, Energy = U at top of swing = m g L (1- cos theta) approx m g L (1 - (1-1/2 theta^2)) = mgL/2 theta^2 for small angles. So, use this expression to answer the question. 

In some problems like this one you need approximations from Taylor series approximations. So, 

For small x: 

(1+x)^P approx 1+px

1-exp(x) approx x

sin x approx x

1-cos x approx x^2/2   

Get used to doing this as a physicist! 

If you have (r+R)^p for R>>r, then first factor as R^p(1+(r/R))^p approx R^p(1+pr/R).