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

September 19, 2014

Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 22

Damped Oscillator

The modified Euler Lagrange equation (including the dissipative term) is m¨ q + 2α˙ q + kq = 0 Define ω0 =

• k

m and γ = α m, the equation for the damped oscillator is ¨ q + 2γ ˙ q + ω2

0q = 0

seek a solution of the form q = q0eλt. The characteristic equation becomes λ2 + 2γλ + ω2

0 = 0 with λ = −γ ±

• γ2 − ω2

The solution is q = e−γt Ae √

γ2−ω2

0t + e−√

γ2−ω2

0t Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -22

September 22, 2014 1 / 19

Underdamped Oscillations

Underdamped Motion: If γ < ω2

0, the motion is underdamped and

the solution is q = q0e−γt cos(Ωt + δ) with Ω =

• ω2

0 − γ2,which is lower than the undamped frequency.

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 > ω0, the motion is exponentially damped q(t) = Ae−γ1t + Be−γ2t with γ1,2 = γ ∓

• γ2 − ω2

0 are both

positive. Critical Damping: If γ = ω0, the two roots of the characteristic equation become equal (λ = −γ = −ω0) and we need to start afresh as a second order differential equation must have two constants. The equation is ¨ q + 2γ ˙ q + ω2

0q = 0. Let us try a solution of type

q(t) = u(t)eλt, 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 eλt + uλeλt d2q dt2 = eλt d2u dt2 + 2λdu dt + λ2

• Substituting these, we get

d2u dt2 + 2(λ + γ)du dt + (λ2 + 2γλ + ω2

0)u = 0

Since λ = −γ = −ω0, the equation become d2u dt2 = 0 which has the solution u(t) = A + Bt. Thus q(t) = (A + Bt)e−γt 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) = −qF0 cos ωt L = 1 2m ˙ q2 − 1 2kq2 + qF(t) Al;ing with Rayleigh dissipation term, the equation of motion becomes ¨ q + 2γ ˙ q + ω2

0q = F0

m eiωt The solution of the homogeneous equation soon goes to equilibrium after the transients (for the unde-damped case) die down. We look

• nly 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−iωt so that ˙ q = −iωq and ¨ q = −ω2q. We get (−ω2 − 2iωγ + ω2

0)Ae−iωt = F0

m e−iωt We get A = F0 m 1 ω2

0 − ω2 − 2iωγ

= F0/m

• (ω2

0 − ω2)2 + 4γ2ω2

e−iϕ with ϕ = tan−1

• 2γω

ω2 − ω2

• Dipan Ghosh (I.I.T. Bombay)
• Class. Mech. -22

September 22, 2014 7 / 19

Driven Oscillations

We get q = ℜ

• Ae−iωt

=| A | cos(ωt + ϕ) | A |= F0/m

• (ω2

0 − ω2)2 + 4γ2ω2

ϕ = tan−1

• 2ωγ

ω2

0 − ω2

• 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 = ω0, 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 = F0/m ω2 = F0 k which corresponds to applying a 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 ω0, 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. As ω → ∞, A = F0 ω2m. 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 3N − k degrees of freedom. we need 3N − k generalized coordinates to describe the system. If, however, there are m non-holonomic constraints, we still have 3N − 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) = x2 8 + y2 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 + λh 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 − λy = 0 x2 8 + y2 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) fmax = 2 Two minimum values at (1, −2) and at (−1, 2) with fmin = −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

• j
• dt

d dt ∂L ∂ ˙ qj

• − ∂L

∂qj

• δqj = 0

Using the fact that the generalized coordinates are independent, we derived the Euler Lagrange equation for each generalized coordinate d dt ∂L ∂ ˙ qj

• − ∂L

∂qj = 0 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

• j=1

ar,jdqj + br,hdt = 0 where ar,j and brj 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 ar,jdqj = 0. n

• j=1
• dt
• d

dt ∂L ∂ ˙ qj

• − ∂L

∂qj +

m

• r=1

λrar,j

• δqj = 0

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. δqj are not independent as they satisfy the constraint condition n

j=1 ar,jδqj = 0. However, since we have m of the λr s are yet

undetermined, we can choose them such that the integrands for j = 1 to m are zero. d dt ∂L ∂ ˙ qj

• − ∂L

∂qj +

m

• r=1

λrar,j = 0 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

n

• j=m+1
• dt
• d

dt ∂L ∂ ˙ qj

• − ∂L

∂qj +

m

• r=1

λrar,j

• δqj = 0

However, these qj s are independent and, therefore for j = m + 1, n d dt ∂L ∂ ˙ qj

• − ∂L

∂qj +

m

• r=1

λrar,j = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -22

September 22, 2014 18 / 19

Euler Lagrange Equations with constraints

We thus have one single form of equation for each of the n generalized coordinates d dt ∂L ∂ ˙ qj

• − ∂L

∂qj +

m

• r=1

λrar,jδqj = 0 Define Generalized Force corresponding to the constraint conditions Qj = − m

j=1 λrar,j.

The Euler Lagrange Equation is given by d dt ∂L ∂ ˙ qj

• − ∂L

∂qj = Qj

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -22

September 22, 2014 19 / 19