EP 222: Classical Mechanics - Lecture 19 Dipan K. Ghosh Indian Institute of Technology Bombay dipan.ghosh@gmail.com Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 19 September 11, 2014
Small Oscillations When a conservative system is slightly displaced from its stable equilibrium position,it undergoes oscillation. Cause of oscillation is restoring force which can do both positive and negative work. When work done is positive, it converts potential energy to kinetic energy and vice versa. For mechanical systems, not far from equilibrium, the restoring force is proportional to displacement ( F = − kx ). For such linear oscillators oscillation frequencies are independent of amplitude Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 1 / 20
Small Oscillations Oscillator motion can be damped in the presence of resistive forces Resistive forces extract energy out of the oscillator For low velocities, the resistive forces are proportional to the velocity. For damped and undamped oscillators, energy can new supplied continuously by an external agency to maintain oscillation. These are called driven oscillations or forced oscillations The amplitude of driven oscillations becomes very large when the driving frequency approaches the natal frequency, a phenomenon known as ”resonance”. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 2 / 20
Small Oscillations Consider a conservative system with { q i } as the generalized coordinates. The forces are derivable from a potential energy function V ( q 1 , q 2 , . . . , q n ). Equilibrium is defined (by Lagrange) as a configuration where all the generalized forces vanish, i.e. ∂ V = 0 ∂ q i In this situation the system will not change its configuration. However, even when Q i = 0, the system may not be stable. If disturbed from an equilibrium position, the system will return to its original configuration only if the equilibrium is stable. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 3 / 20
Equilibrium The potential energy is V ( θ ) = + mgl (1 − cos θ ). The generalized force corresponding to θ is Q θ = − ∂ V ∂θ = − mgl sin θ = − mgx The generalized ”force” is actually a restoring torque. Q θ = 0 for θ = 0 , π . L = 1 2 ml 2 ˙ θ 2 − mgl (1 − cos θ ) Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 4 / 20
Equilibrium For θ = 0, L = 1 θ 2 − 1 2 mgl θ 2 , so that V ( θ ) = +1 2 ml 2 ˙ 2 mgl θ 2 Q θ = − mgl θ , which is of restoring nature. Consider the equilibrium at θ = π , Near this position cos θ = cos( π + δθ ) = − cos δθ ≈ − 1 + 1 2 δθ 2 In this situation L = 1 θ 2 + 1 2 ml 2 ˙ 2 mgl ( δθ ) 2 , the corresponding force is anti-restoring. ∂ 2 V For stable equilibrium we have > 0 ∂ q i ∂ q j Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 5 / 20
Equilibrium Without loss of generality, let us shift the equilibrium position to the origin ( q 1 = q 2 = . . . = 0). If this system is disturbed to a position { q i } , we can write � ∂ 2 V � ∂ V � � q i +1 � � V ( q 1 , q 2 , . . . ) = V (0 , 0 , . . . )+ q i q i + . . . ∂ q i 2 ∂ q i ∂ q j 0 0 i i , j Measure potential energy from the origin. At equilibrium, the second term is zero. Neglect higher order terms. For stable equilibrium � ∂ 2 V � 1 � > 0 2 ∂ q i ∂ q j 0 i , j � ∂ 2 V � Let V ij = ∂ q i ∂ q j 0 V = 1 � i , j V i , j q i q j , with V ij = V ji 2 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 6 / 20
Equilibrium For a sceloronomic system, the kinetic energy, in general, T = 1 � i , j t ij ˙ q i ˙ q j , where the coefficients T i , j could be functions of 2 generalized coordinates. However, it turns out that the dependence on generalized coordinates is weak and we can essentially treat them as constants. Around the equilibrium, we have L = T − V = 1 � ( t ij ˙ q i ˙ q j − V ij q i q j ) 2 i , j � ∂ T Use Euler Lagrange Equation d � − ∂ V = 0 dt ∂ ˙ q k ∂ q k Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 7 / 20
Small Oscillations Euler Lagrange equation gives d 1 q i δ kj ] + 1 � � 2[ t ij δ ik ˙ q j + t ij ˙ V ij ( δ ik q j + q i δ k , j ) = 0 2 dt i , j i , j 1 + 1 � � � � = 0 t kj ¨ q j + t ik ¨ V ik q j + q i V ik q i 2 2 j i j i Change the dummy index from j to i � � t ik ¨ q i + V ik q i = 0 i i Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 8 / 20
Characteristic Frequencies We seek a solution of the form q i = A i e i ω t i ( V ik − ω 2 t ik ) A i = 0 � Solution exists if det( V − ω 2 T ) = 0 This is a single equation of n − th degree in ω 2 . It has n roots, some of which are real some complex. The real toots of the equation derived from above are called Characteristic frequencies or Eigenfrequencies For real physical situations, the roots are real and positive. For if ω had an imaginary part, the energy would not be conserved. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 9 / 20
Characteristic Frequencies i , k ( V ik − ω 2 t ik ) A ∗ We have � k A i = 0, so that � ik V ik A ∗ k A i ω 2 = � i , k t ik A ∗ k A i Since V ik = V ki and t ik = t ki both numerator and denominator are real. They are also positive because if we write A i = a i + ib i , � � t ik A ∗ i A k = T ik ( a i − ib i )( a k + ib k ) i , k i , k � = t ik ( a i a k + b i b k ) i , k The imaginary terms cancel because of symmetry of t ik . The right hand side is a sum of two positive definite terms ˜ aTa Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 10 / 20
Matrix Formulation k ( V ki − λ t ki ) A i = 0, where λ = ω 2 We have � Define A 1 V 11 V 12 . . . V 1 N . . . A 2 V 21 V 22 V 2 N A = V = . . . . . . . . . . . . . . . . . . A N V N 1 V N 2 V NN along with . . . T 11 T 12 T 1 N T 21 T 22 . . . T 2 N T = . . . . . . . . . . . . T N 1 T N 2 . . . T NN We then have the matrix equation VA = λ TA This is not an eigenvalue equation as VA is not proportional to A . If T is invertible, we can convert this into an eigenvalue equation T − 1 VA = λ IA Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 11 / 20
Matrix Formulation This equation has N solutions. Each solution is called a MODE . Let the k − th mode frequency be λ k = ω 2 k . Let the corresponding vector A be written as A k 1 A k 2 A k = . . . A kN We have VA k = λ k TA k (1) Take conjugate of this equation and change the index k to i , A i V = λ i ˜ ˜ A i T (2) From (1) we get ˜ A k VA k λ k = ˜ A k TA k Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 12 / 20
Matrix Formulation From (1) and (2), we get ( λ k − λ i )˜ A i TA k = 0 If the eigenvalues are non-degenerate, this gives the ”orthogonallty condition” (different from usual) ˜ A i TA k = 0 We define normalization condition ˜ A i TA i = 1 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 13 / 20
Normal Modes - Example Let the generalized coordinates be displacement of the masses from their equilibrium positions. Let m 1 be displaced by x 1 and m 2 by x 2 . The central spring is then compressed or extended by x 2 − x 1 . Traditional method of solution: m 1 ¨ x 1 = − 4 kx 1 − k ( x 1 − x 2 ) m 2 ¨ x 2 = − 2 kx 2 − k ( x 2 − x 1 ) Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 14 / 20
Normal Modes - Example - Traditional Solution We can rewrite the pair of equations as x 2 ) = − 7 m (¨ x − 1 − ¨ 2 k ( x 1 − x 2 ) m (2¨ x 2 + ¨ x 1 ) = − 2 k (2 x 1 + x 2 ) Define Normal Coordinates y 1 = x 1 − x 2 y 2 = 2 x 1 + x 2 The equations are decoupled y 1 = − 7 m ¨ 2 ky 1 m ¨ y 2 = − 2 ky 2 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 15 / 20
Normal Modes - Example � The normal coordinates oscillate with frequencies 2 k / m and � 7 k / 2 m , which are the normal mode frequencyes. y α + ω 2 The new normal coordinates satisfy ¨ α y α = 0 The Lagrangian of the system is L = 1 1 + 1 2 − 1 1 − 1 2 k 2 ( x 2 − x 1 ) 2 − 1 x 2 x 2 2 k 1 x 2 2 k 3 x 2 2 m ˙ 2 m 2 ˙ 2 Define t and V matrices ∂ 2 T � m 1 � 0 t = = ∂ ˙ x i ∂ ˙ 0 m 2 x j ∂ 2 V � 5 k � − k V = = − k 3 k ∂ ˙ x i ∂ ˙ x j Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 16 / 20
Normal Modes - Example Condition for existence of non-trivial solution det( V ik − ω 2 t ik ) = 0, which gives � 5 k − 2 m ω 2 � − k � � � = 0 � 3 k − m ω 2 � − k � Solution ω 2 = 7 k 2 m and 2 k m Finding the normal modes: ω 2 TA = VTA � A 1 � Let A = A 2 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 17 / 20
Normal Modes - Example For ω 2 = 7 k 2 m � 5 k − k � � A 1 � = 7 k � 2 m 0 � � A 1 � − k 3 k A 2 0 m A 2 2 m This gives a − 2 = − 2 A 1 , so that � 1 � A = − 1 � 1 � For ω 2 = 2 k m , we have 1 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -19 September 18, 2014 18 / 20
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