Goldstein Solution Chapter 6 prob 13 pg 274

In[1611]:= "Programmer: Hemanta Bhattarai""Institution: CDP, TU""Problem: Classical Mechanics ,Goldstein, Third Edition, Chapter 6 Problem 13 pg.274""Date: 2013 Feb 8"ClearAll@k, l, qDl = a;q = q;m@1D = m; m@2D = m;k = k;H*The Kinetic Energy of the system where m@D are masses and x@D are extensions*LT = 1 2 Hm@1D x@1D^2 + m@2D x@2D^2L;H*Potential energy of the system including electrostatic potential energy*LV = 1 2 k Hx@1D^2 + Hx@2D - x@1DL^2 + x@2D^2L + q^2 Hx@2D - x@1D + lL;H*Tij matrix*LTij = Table@D@T, x@iD, x@jDD, 8i, 1, 2<, 8j, 1, 2<D;H*Vij matrix*LVij = Table@D@V, x@iD, x@jDD, 8i, 1, 2<, 8j, 1, 2<D;H*Secular matrix with Tij and Vij determined at initial equlibrium position*Lfx = W * Tij - Vij . 8Hx@2D - x@1DL ® 0<;

d = Det@fxD;H*frequencies of oscillation*Lomga = Solve@d 0, WD Simplify;

g1 = fx . omga@@1DD;g2 = fx . omga@@2DD;

H*Determining the value of a@D's for each mode of oscillation*Leq1 = Table@g1@@iDD.8a@1D, a@2D<, 8i, 1, 2<D;sol1 = Solve@8eq1@@1DD 0, eq1@@2DD 0, m@1D a@1D^2 + m@2D a@2D^2 1<, 8a@1D, a@2D<D;

eq2 = Table@g2@@iDD.8a@1D, a@2D<, 8i, 1, 2<D;

sol2 = Solve@8eq2@@1DD 0, eq2@@2DD 0, m@1D a@1D^2 + m@2D a@2D^2 1<, 8a@1D, a@2D<D;"ððððððððððððððððððððððððððððððððððððððððððððððððððððððððððð""**************************************************************""The KE term is"T"The PE term is"V

"The secular equation is: where W=Ω^2"d 0"The square of modes of frequency is i.e Ω^2"omga N"The first set of a's for first frequency "sol1@@2DD N"The second a's are for second frequency "sol2@@2DD N

"***************************************************************""ððððððððððððððððððððððððððððððððððððððððððððððððððððððððððð"

Out[1611]= Programmer: Hemanta Bhattarai

Out[1612]= Institution: CDP, TU

Out[1613]= Problem: Classical Mechanics ,Goldstein, Third Edition, Chapter 6 Problem 13 pg.274

Out[1614]= Date: 2013 Feb 8

Out[1633]= ððððððððððððððððððððððððððððððððððððððððððððððððððððððððððð

Out[1634]= **************************************************************

Out[1635]= The KE term is

Out[1636]=1

2Im x@1D2

+ m x@2D2M

Out[1637]= The PE term is

Out[1638]=q2

a - x@1D + x@2D+1

2k Ix@1D2

+ x@2D2+ H-x@1D + x@2DL2M

Out[1639]= The secular equation is: where W=Ω^2

Out[1640]= 3 k2 +4 k q2

a3- 4 k m W -

4 m q2 W

a3+ m2 W2 0

Out[1641]= The square of modes of frequency is i.e Ω^2

Out[1642]= ::W ®k

m>, :W ®

3. a3 k + 4. q2

a3 m>>

Out[1643]= The first set of a's for first frequency

Out[1644]= :[email protected] ®0.707107

m, [email protected] ®

0.707107

m>

Out[1645]= The second a's are for second frequency

Out[1646]= :[email protected] ®0.707107

m, [email protected] ® -

0.707107

m>

Out[1647]= ***************************************************************

Out[1648]= ððððððððððððððððððððððððððððððððððððððððððððððððððððððððððð

2 three-mass-spring-goldstein-with_charge.txt.nb

Goldstein Solution Chapter 6 prob 13 pg 274

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