EP 222: Classical Mechanics - Lecture 19 Dipan K. Ghosh Indian - - PowerPoint PPT Presentation

EP 222: Classical Mechanics - Lecture 19

Dipan K. Ghosh

Indian Institute of Technology Bombay dipan.ghosh@gmail.com

September 11, 2014

Dipan K. Ghosh EP 222: Classical Mechanics - Lecture 19

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 {qi} as the generalized coordinates. The forces are derivable from a potential energy function V (q1, q2, . . . , qn). Equilibrium is defined (by Lagrange) as a configuration where all the generalized forces vanish, i.e. ∂V ∂qi = 0 In this situation the system will not change its configuration. However, even when Qi = 0, the system may not be stable. If disturbed from an equilibrium position, the system will return to its

• riginal 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 Qθ = −∂V ∂θ = −mgl sin θ = −mgx The generalized ”force” is actually a restoring torque. Qθ = 0 for θ = 0, π. L = 1 2ml2 ˙ θ2 − mgl(1 − cos θ)

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -19

September 18, 2014 4 / 20

Equilibrium

For θ = 0, L = 1 2ml2 ˙ θ2 − 1 2mglθ2, so that V (θ) = +1 2mglθ2 Qθ = −mglθ, which is of restoring nature. Consider the equilibrium at θ = π, Near this position cos θ = cos(π + δθ) = − cos δθ ≈ −1 + 1 2δθ2 In this situation L = 1 2ml2 ˙ θ2 + 1 2mgl(δθ)2, the corresponding force is anti-restoring. For stable equilibrium we have ∂2V ∂qi∂qj > 0

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

• rigin (q1 = q2 = . . . = 0). If this system is disturbed to a position

{qi}, we can write V (q1, q2, . . .) = V (0, 0, . . .)+

• i

∂V ∂qi

• qi+1

2

• i,j

∂2V ∂qi∂qj

• qiqi+. . .

Measure potential energy from the origin. At equilibrium, the second term is zero. Neglect higher order terms. For stable equilibrium 1 2

• i,j

∂2V ∂qi∂qj

• > 0

Let Vij = ∂2V ∂qi∂qj

• V = 1

2

• i,j Vi,jqiqj, with Vij = Vji

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 2

• i,j tij ˙

qi ˙ qj, where the coefficients Ti,j could be functions of generalized coordinates. However, it turns out that the dependence

• n generalized coordinates is weak and we can essentially treat them

as constants. Around the equilibrium, we have L = T − V = 1 2

• i,j

(tij ˙ qi ˙ qj − Vijqiqj) Use Euler Lagrange Equation d dt ∂T ∂ ˙ qk

• − ∂V

∂qk = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -19

September 18, 2014 7 / 20

Small Oscillations

Euler Lagrange equation gives d dt

• i,j

1 2[tijδik ˙ qj + tij ˙ qiδkj] + 1 2

• i,j

Vij(δikqj + qiδk,j) = 0 1 2  

j

tkj¨ qj +

• i

tik¨ qi   + 1 2  

j

Vikqj +

• i

Vikqi   = 0 Change the dummy index from j to i

• i

tik¨ qi +

• i

Vikqi = 0

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -19

September 18, 2014 8 / 20

Characteristic Frequencies

We seek a solution of the form qi = Aieiωt

• i(Vik − ω2tik)Ai = 0

Solution exists if det(V − ω2T) = 0 This is a single equation of n−th degree in ω2. It has n roots, some

• f 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

We have

i,k(Vik − ω2tik)A∗ kAi = 0, so that

ω2 =

• ik VikA∗

kAi

• i,k tikA∗

kAi

Since Vik = Vki and tik = tki both numerator and denominator are real. They are also positive because if we write Ai = ai + ibi,

• i,k

tikA∗

i Ak =

• i,k

Tik(ai − ibi)(ak + ibk) =

• i,k

tik(aiak + bibk) The imaginary terms cancel because of symmetry of tik. 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

We have

k(Vki − λtki)Ai = 0, where λ = ω2

Define A =     A1 A2 . . . AN     V =     V11 V12 . . . V1N V21 V22 . . . V2N . . . . . . . . . . . . VN1 VN2 . . . VNN     along with T =     T11 T12 . . . T1N T21 T22 . . . T2N . . . . . . . . . . . . TN1 TN2 . . . TNN     We then have the matrix equation VA = λTA 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 −1VA = λIA

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 λk = ω2

• k. Let the corresponding vector

A be written as Ak =     Ak1 Ak2 . . . AkN     We have VAk = λkTAk (1) Take conjugate of this equation and change the index k to i, ˜ AiV = λi ˜ AiT (2) From (1) we get λk = ˜ AkVAk ˜ AkTAk

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -19

September 18, 2014 12 / 20

Matrix Formulation

From (1) and (2), we get (λk − λi)˜ AiTAk = 0 If the eigenvalues are non-degenerate, this gives the ”orthogonallty condition” (different from usual) ˜ AiTAk = 0 We define normalization condition ˜ AiTAi = 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 m1 be displaced by x1 and m2 by x2. The central spring is then compressed or extended by x2 − x1. Traditional method of solution: m1¨ x1 = −4kx1 − k(x1 − x2) m2¨ x2 = −2kx2 − k(x2 − x1)

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 m(¨ x − 1 − ¨ x2) = −7 2k(x1 − x2) m(2¨ x2 + ¨ x1) = −2k(2x1 + x2) Define Normal Coordinates y1 = x1 − x2 y2 = 2x1 + x2 The equations are decoupled m¨ y1 = −7 2ky1 m¨ y2 = −2ky2

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -19

September 18, 2014 15 / 20

Normal Modes - Example

The normal coordinates oscillate with frequencies

• 2k/m and
• 7k/2m, which are the normal mode frequencyes.

The new normal coordinates satisfy ¨ yα + ω2

αyα = 0

The Lagrangian of the system is L = 1 2m ˙ x2

1 + 1

2m2 ˙ x2

2 − 1

2k1x2

1 − 1

2k2(x2 − x1)2 − 1 2k3x2

2

Define t and V matrices t = ∂2T ∂ ˙ xi∂ ˙ xj = m1 m2

• V =

∂2V ∂ ˙ xi∂ ˙ xj = 5k −k −k 3k

• 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(Vik − ω2tik) = 0, which gives

• 5k − 2mω2

−k −k 3k − mω2

• = 0

Solution ω2 = 7k 2m and 2k m Finding the normal modes: ω2TA = VTA Let A = A1 A2

• Dipan Ghosh (I.I.T. Bombay)
• Class. Mech. -19

September 18, 2014 17 / 20

Normal Modes - Example

For ω2 = 7k 2m 5k −k −k 3k A1 A2

• = 7k

2m 2m m A1 A2

• This gives a − 2 = −2A1, so that

A = 1 −1

• For ω2 = 2k

m , we have 1 1

• Dipan Ghosh (I.I.T. Bombay)
• Class. Mech. -19

September 18, 2014 18 / 20

Normal Modes - Example

The general solution can be written as x1 x2

• =

1 1

• cos ω−t+B

1 1

• sin ω−t+C

1 −2

• cos ω+t+D

1 −2

• sin ω+t

Suppose we pull the two masses to one side (or one to each side) and

• release. We have x1(0) = x10, x2(0) = x20; ˙

x1(0) = ˙ x2(0) = 0, We have, A + C = x10, A − 2C = x20, Bω− + Dω+ = 0, Bω− − 2Dω+ = 0 The solutions are x1 = 2 3x10 + 1 3x20

• cos ω−t +

1 3x10 − 1 3x20

• cos ω+t

x2 = 2 3x10 + 1 3x20

• cos ω−t −

2 3x10 − 2 3x20

• cos ω+t

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -19

September 18, 2014 19 / 20

Normal Modes - Example

The solutions can be written as x1 − x2 = (x10 − x20) cos ω+t 2x1 + x2 = (2x10 + x20) cos ω−t

Dipan Ghosh (I.I.T. Bombay)

• Class. Mech. -19

September 18, 2014 20 / 20