EP 222: Classical Mechanics - Lecture 22 Dipan K. Ghosh Indian - PowerPoint PPT Presentation

EP 222: Classical Mechanics - Lecture 22 Dipan K. Ghosh Indian Institute of Technology Bombay dipan.ghosh@gmail.com Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 22 September 19, 2014

Damped Oscillator The modified Euler Lagrange equation (including the dissipative term) is m ¨ q + 2 α ˙ q + kq = 0 � m and γ = α k Define ω 0 = m , the equation for the damped oscillator is q + ω 2 ¨ q + 2 γ ˙ 0 q = 0 seek a solution of the form q = q 0 e λ t . The characteristic equation becomes λ 2 + 2 γλ + ω 2 γ 2 − ω 2 � 0 = 0 with λ = − γ ± The solution is √ 0 t + e − √ q = e − γ t � 0 t � γ 2 − ω 2 γ 2 − ω 2 Ae Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 1 / 19

Underdamped Oscillations Underdamped Motion: If γ < ω 2 0 , the motion is underdamped and the solution is q = q 0 e − γ t cos(Ω t + δ ) � ω 2 0 − γ 2 ,which is lower than the undamped frequency. with Ω = The amplitude of oscillation decreases with time exponentially. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 2 / 19

Overdamped and Critically Damped Motion Overdamped Moption: If γ > ω 0 , the motion is exponentially � damped q ( t ) = Ae − γ 1 t + Be − γ 2 t with γ 1 , 2 = γ ∓ γ 2 − ω 2 0 are both positive. Critical Damping: If γ = ω 0 , the two roots of the characteristic equation become equal ( λ = − γ = − ω 0 ) and we need to start afresh as a second order differential equation must have two constants. q + ω 2 The equation is ¨ q + 2 γ ˙ 0 q = 0. Let us try a solution of type q ( t ) = u ( t ) e λ t , where λ is the solution of the characteristic equation and in this case λ = − γ Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 3 / 19

Critical Damping dq dt = du dt e λ t + u λ e λ t d 2 q � d 2 u � dt 2 + 2 λ du dt + λ 2 dt 2 = e λ t Substituting these, we get d 2 u dt 2 + 2( λ + γ ) du dt + ( λ 2 + 2 γλ + ω 2 0 ) u = 0 Since λ = − γ = − ω 0 , the equation become d 2 u dt 2 = 0 which has the solution u ( t ) = A + Bt . Thus q ( t ) = ( A + Bt ) e − γ t Approach to equilibrium is the fastest in this case. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 4 / 19

Damping- Approach to Equilibrium Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 5 / 19

Driven Oscillations In addition to the harmonic potential and the damping force, let there be a sinusoidal potential − qF ( t ) = − qF 0 cos ω t L = 1 q 2 − 1 2 kq 2 + qF ( t ) 2 m ˙ Al;ing with Rayleigh dissipation term, the equation of motion becomes 0 q = F 0 q + ω 2 m e i ω t q + 2 γ ˙ ¨ The solution of the homogeneous equation soon goes to equilibrium after the transients (for the unde-damped case) die down. We look only for the particular solution. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 6 / 19

Driven Oscillations The equation is solved by an ansatz q = Ae − i ω t so that ˙ q = − i ω q q = − ω 2 q . and ¨ We get 0 ) Ae − i ω t = F 0 ( − ω 2 − 2 i ωγ + ω 2 m e − i ω t We get 1 A = F 0 0 − ω 2 − 2 i ωγ ω 2 m F 0 / m e − i ϕ = � 0 − ω 2 ) 2 + 4 γ 2 ω 2 ( ω 2 with � 2 γω � ϕ = tan − 1 ω 2 − ω 2 0 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 7 / 19

Driven Oscillations � Ae − i ω t � We get q = ℜ = | A | cos( ω t + ϕ ) F 0 / m | A | = � 0 − ω 2 ) 2 + 4 γ 2 ω 2 ( ω 2 � � 2 ωγ ϕ = tan − 1 ω 2 0 − ω 2 Motion is oscillatory with the frequency of oscillation the same as that of the driving frequency. The amplitude of oscillation has a peak at ω = ω 0 , the natural frequency and the height of peak increases with decrease of the damping strength γ . Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 8 / 19

Driven Oscillations- Resonance At zero frequency A = F 0 / m = F 0 k which corresponds to applying a ω 2 0 const ant force to the mass and stretch it. The amount of stretching then only depends on the applied force and the spring constant. The mass and the drag are irrelevant. At low frequencies the argument of tan − 1 remains positive and increases from zero to π/ 2 as the frequency approaches the resonant frequency. As ω exceeds ω 0 , the argument of arc-tan becomes negative and the phase angle becomes greater than π/ 2 approaching π at very large frequencies so that the response of the oscillator becomes anti-phase with the driving frequency. F 0 As ω → ∞ , A = ω 2 m . If we are pushing the system at very high frequency, the stiffness of the spring or the drag becomes irrelevant and only the inertia becomes important. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 9 / 19

Driven Oscillations- Resonance Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 10 / 19

Lagrange Multipliers For N particles with k holonomic constraints, there are 3 N − k degrees of freedom. we need 3 N − k generalized coordinates to describe the system. If, however, there are m non-holonomic constraints, we still have 3 N − k generalized coordinates. These additional m quantities are removed from the problem by method of Lagrange Multipliers. Consider the problem of minimisation of f ( x , y ) = xy subject to constraint h ( x , y ) = x 2 8 + y 2 2 − 1 = 0 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 11 / 19

Lagrange Multipliers The level curves for f ( x , y ) = xy are hyperbolas. Suppose the level curve of f ( x , y ) intersects the constraint curve (ellipse) at P. If we move along the constraint curve to the right of P, the value of xy increases whereas if we move to the left of P, the value of xy decreases. Thus P cannot be an optimal point (min/max). At the optimal point the level curve and the constraint curve must have a common tangent. Thus the set of points S which satisfies h ( x , y ) = 0 and ∇ f + λ ∇ h = 0 for some λ contains the extremal value of f subject to given constraint. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 12 / 19

Lagrange Multipliers Setting the gradient of f + λ h to zero, we have ∂ f ∂ x + λ∂ h ∂ x = 0 ∂ f ∂ y + λ∂ h ∂ y = 0 h = 0 For the present problem, we have y − λ 4 x = 0 x − λ y = 0 x 2 8 + y 2 2 − 1 = 0 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 13 / 19

Lagrange Multipliers Solution gives λ = ± 2. Substituting this value of λ into the equations and solving we get, Two maximum values of f = xy at (1 , 2) and at ( − 1 , − 2) f max = 2 Two minimum values at (1 , − 2) and at ( − 1 , 2) with f min = − 2 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 14 / 19

Euler Lagrange Equations with constraints The Euler Lagrange equations were derived from the optimistion of the action integral � d � ∂ L � � � − ∂ L � dt δ q j = 0 dt ∂ ˙ q j ∂ q j j Using the fact that the generalized coordinates are independent, we derived the Euler Lagrange equation for each generalized coordinate � ∂ L � d − ∂ L = 0 ∂ ˙ ∂ q j dt q j In case of non-holonomic constraints, the generalized coordinates are not independent (this can also happen if the set of generalized coordinates is not chosen to be independent). Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 15 / 19

Euler Lagrange Equations with constraints Suppose there are m number of non-holonomic constraints of type involving generalized coordinates in differential form n � a r , j dq j + b r , h dt = 0 j =1 where a r , j and b rj may depend on j and t . r is an index which runs from 1 to m . This is actually m equations, one equation for each value of r . Since virtual displacements take over constant time, we can vary the path from actual motion by replacing the constraints by � n j =1 a r , j dq j = 0. � ∂ L n � m � � � d − ∂ L � � dt + λ r a r , j δ q j = 0 ∂ ˙ ∂ q j dt q j j =1 r =1 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 16 / 19

Euler Lagrange Equations with constraints The additional term is actually zero because for each r = 1 , m , the sum over j is zero. δ q j are not independent as they satisfy the constraint condition � n j =1 a r , j δ q j = 0. However, since we have m of the λ r s are yet undetermined, we can choose them such that the integrands for j = 1 to m are zero. � ∂ L m � d − ∂ L � + λ r a r , j = 0 ∂ ˙ ∂ q j dt q j r =1 Note that the last term is not zero because the sum over j is missing. Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 17 / 19

Euler Lagrange Equations with constraints This leaves us with � ∂ L n � m � � � d − ∂ L � � dt + λ r a r , j δ q j = 0 dt ∂ ˙ q j ∂ q j r =1 j = m +1 However, these q j s are independent and, therefore for j = m + 1 , n � ∂ L m � d − ∂ L � + λ r a r , j = 0 ∂ ˙ ∂ q j dt q j r =1 Dipan Ghosh (I.I.T. Bombay) Class. Mech. -22 September 22, 2014 18 / 19