Classical Mechanics

Oscillations and Waves

Quantum and Thermal Physics

PHY293: Modern Physics Section 5.1: Schrodinger’s Wave Equation Sunday, 20 November, 11 Wave-particle duality Sunday, 20 November, 11 Wave-particle duality  Tonomura (Hitachi) experiment Sunday, 20 November, 11 Wave-particle duality  Tonomura (Hitachi) experiment  Wave nature of electron Sunday, 20 November, 11 Wave-particle duality  Tonomura (Hitachi) experiment  Wave nature of electron  Single electron interference Sunday, 20 November, 11 Wave-particle duality  Tonomura (Hitachi) experiment  Wave nature of electron  Single electron interference  Shows complementary nature of wave-particle Sunday, 20 November, 11 Wave-particle duality  Tonomura (Hitachi) experiment  Wave nature of electron  Single electron interference  Shows complementary nature of wave-particle Sunday, 20 November, 11 Wave-particle duality  Tonomura (Hitachi) experiment  Wave nature of electron  Single electron interference  Shows complementary nature of wave-particle Sunday, 20 November, 11 Wave-particle duality  Tonomura (Hitachi) experiment  Wave nature of electron  Single electron interference  Shows complementary nature of wave-particle  What is the physical quantity that is responsible for the interference? Sunday, 20 November, 11 Wave-particle duality  What is the physical quantity that is responsible for the interference? Sunday, 20 November, 11 Wave-particle duality  What is the physical quantity that is responsible for the interference?  Analogy with E+M wave: Sunday, 20 November, 11 Wave-particle duality  What is the physical quantity that is responsible for the interference?  Analogy with E+M wave: Sunday, 20 November, 11 Wave-particle duality  What is the physical quantity that is responsible for the interference?  Analogy with E+M wave: Sunday, 20 November, 11 Schrondinger Sunday, 20 November, 11 Schrondinger  Master’s Thesis in Experimental Physics Sunday, 20 November, 11 Schrondinger  Master’s Thesis in Experimental Physics  Almost leaves physics to get married (Couldn’t get approved by fiance’s mother due to his low salary) Sunday, 20 November, 11 Schrondinger  Master’s Thesis in Experimental Physics  Almost leaves physics to get married (Couldn’t get approved by fiance’s mother due to his low salary)  Temporarily Leaves physics for WWI <though still productive> Sunday, 20 November, 11

PHY | 293 | Oscillations and Waves | study guides | PHY293: Modern Physics Section

Suggested Problems # 3 Solutions QUESTION 1 - The triatomic molecule problem We look at an idealized version of a triatomic molecule, with a larger mass M at the center, and two smaller masses m on each side, connected by springs of stiffness k , which represent the binding force between the atoms. We consider simple harmonic oscillations between the atoms, of the form η j = η 0 j cos ( ωt ) . Part A: EOM and normal frequencies Using a free-body diagram of the situation, it is possible to note that the force felt by the mass on the left (mass 1) will be proportional to the displacement of mass 2, minus the displacement of mass 1. For mass 2, the force will depend on the extension of the springs on both sides, whereas for mass 3, the situation will be very similar to mass 1, except for the fact the spring is connected to the left of the mass. This is represented mathematically as: m ¨ η 1 =( η 2 − η 1 ) k (1) M ¨ η 2 = − ( η 2 − η 1 ) k +( η 3 − η 2 )=( η 1 + η 3 − 2 η 2 ) k (2) m ¨ η 3 = − ( η 3 − η 2 ) k (3) Since η j = η 0 j cos ( ωt ) , we ﬁnd that ¨ η j = − ω 2 η j . Replacing these in the above equations, and rearranging: − mω 2 η 0 1 cos ( ωt )+ � η 0 1 cos ( ωt ) − η 0 2 cos ( ωt ) � k =0 (4) − Mω 2 η 0 2 cos ( ωt )+ � 2 η 0 2 cos ( ωt ) − η 0 1 cos ( ωt ) − η 0 3 cos ( ωt ) � k =0 (5) − mω 2 η 0 3 cos ( ωt )+ � η 0 3 cos ( ωt ) − η 0 2 cos ( ωt ) � k =0 (6) Dividing by cos ( ωt ) and by the respective masses m and M , − ω 2 η 0 1 + � η 0 1 − η 0 2 � k m =0 (7) − ω 2 η 0 2 + � 2 η 0 2 − η 0 1 − η 0 3 � k M =0 (8) − ω 2 η 0 3 + � η 0 3 − η 0 2 � k m =0 (9) In matrix form, this gives:      k m − ω 2 − k m 0 − k M 2 k M − ω 2 − k M 0 − k m k m − ω 2           η 1 η 2 η 3      =0 (10) For a non-trivial solution, we look at situations where the determinant of the 3x3 matrix will be equal to 0. The determinant of an arbitrary square matrix is given by the Leibniz formula, which for a 3x3 matrix A ij takes the form: det ( A ij )= � � � � � � � � � a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 � � � � � � � � � = a 1 1( a 22 a 33 − a 32 a 23 )+ a 21 ( a 32 a 13 − a 12 a 33 )+ a 31 ( a 12 a 23 − a 22 a 13 ) (11) In the case of our 3x3 matrix above, we look at: � � � � � � � � � k m − ω 2 − k m 0 − k M 2 k M − ω 2 − k M 0 − k m k m − ω 2 � � � � � � � � � =0 (12) This is, based on equation (11): � k m − ω 2 � 2 � 2 k M − ω 2 � − 2 k 2 Mm � k m − ω 2 � =0 (13) 2 Rearranging this equation, we obtain: � ω 2 − k m � � � ω 2 � 2 − � k m − 2 k M � ω 2 � =0 (14) � ω 2 − k m � � ω 2 − k ( M +2 m ) mM � ω 2 =0 (15) It is thus easy to show, from equation (15) that the three solutions will be, as presented in the problem statement: ω 2 1 =0 (16) ω 2 2 = k m (17) ω 2 3 = k ( M +2 m ) mM (18) Part B: Normal modes and atom motion CASE 1: ω 2 1 =0 We plug ω 1 =0 into equation (40), which becomes      k m − k m 0 − k M 2 k M − k M 0 − k m k m           η 1 η 2 η 3      =0 (19) Solving this matrix, we obtain the following three equations: k m η 1 − k m η 2 =0 (20) − k M η 1 + 2 k M η 2 − k M η 3 =0 (21) − k m η 2 + k m η 3 =0 (22) 3

PHY | 293 | Oscillations and Waves | assessment | Suggested Problems # 3 Solutio

PHY | 293 | Oscillations and Waves | study guides | 

phy293_l061.pdf-PHY293 Oscillations Lecture #6 September 21,

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