Goldstein Solution Chapter 6 prob 13 pg 274
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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, qD l = 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*L T = 1 2 Hm@1D x@1D ^2 + m@2D x@2D ^2L; H*Potential energy of the system including electrostatic potential energy*L V = 1 2k Hx@1D ^2 + Hx@2D - x@1DL ^2 + x@2D ^2L + q^2 Hx@2D - x@1D + lL; H*Tij matrix*L Tij = Table@D@T, x@iD,x@jDD, 8i, 1, 2<, 8j, 1, 2<D; H*Vij matrix*L Vij = 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*L fx =W* Tij - Vij . 8Hx@2D - x@1DL fi 0<; d = Det@fxD; H*frequencies of oscillation*L omga = 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*L eq1 = 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; "" "**************************************************************" "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 "***************************************************************" "" 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]= Out[1634]= **************************************************************
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Mathematica for the solution of the Goldstein problem on oscilllations
Transcript of 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
