Lecture d31. Linear harmonic oscillator Spring-Mass system k n Spring force F k>0 Newton's second Law m2+k:x=0 (Define) Natural frequency(and period) k 2丌 Equation of a linear harmonic oscillator +wnx=o
Lecture D31 : Linear Harmonic Oscillator Spring-Mass System Spring Force F = −kx, k > 0 Newton’s Second Law mx¨ + kx = 0 (Define) Natural frequency (and period) ωn = s k m τ = 2π ωn Equation of a linear harmonic oscillator x¨ + ω 2 nx = 0 1
Solution General solution c(t=A cos wnt+B sin wnt or xc(t)=Csin(unt+φ) Initial conditions x(0)=x0x(0)=io Solution w(t=o cos wnt+-sin wnt or 1r0 a(t)=va5+(co/wn)sin(wnt +tan( ))
Solution General solution x(t) = A cos ωnt + B sin ωnt or, x(t) = C sin(ωnt + φ) Initial conditions x(0) = x0 x˙(0) = ˙x0 Solution, x(t) = x0 cos ωnt + x˙0 ωn sin ωnt or, x(t) = q x 2 0 + ( ˙x0/ωn) 2 sin(ωnt + tan−1 ( x0ωn x˙0 )) 2
Graphical Representation 2丌 T B t Displacement, velocity and Acceleration 0 2丌
Graphical Representation Displacement, Velocity and Acceleration 3
Energy Conservation st Equilibrium Position No dissipation T+V= constant otential Energy k(a+sst) k 7709 At Equilibrium -kost f mg=0 V=-kr 2
Energy Conservation Equilibrium Position No dissipation T + V = constant Potential Energy V = 1 2 k(x + δst) 2 − 1 2 kδ2 st − mgx At Equilibrium −kδst + mg = 0, V = 1 2 kx2 4
Energy Conservation(cont'd) Kinetic Energy Conservation of energy (T+V)=mii+ kai=0 Governing equation mi+k= o Above represents a very general way of de- riving equations of motion (Lagrangian Me- chanics
Energy Conservation (cont’d) Kinetic Energy 1 2 mx˙ 2 Conservation of energy d dt (T + V ) = mx˙x¨ + kxx˙ = 0 Governing equation mx¨ + kx = 0 Above represents a very general way of deriving equations of motion (Lagrangian Mechanics) 5
Energy Conservation(cont'd) If v=o at the equilibrium position T=Tmo for V≡Vmar T=0 for -max mac vmaz
Energy Conservation (cont’d) If V = 0 at the equilibrium position, V = 0 T = Tmax for x = 0 V = Vmax T = 0 for x = xmax → Tmax = Vmax 6
EXamples o Spring-mass systems Rotating machinery Pendulums(small amplitude) Oscillating bodies(small amplitude) e Aircraft motion( Phugoid) Waves(String, Surface, Volume, etc.) ● Circuits
Examples • Spring-mass systems • Rotating machinery • Pendulums (small amplitude) • Oscillating bodies (small amplitude) • Aircraft motion (Phugoid) • Waves (String, Surface, Volume, etc.) • Circuits • . . . 7
The Phugoid Idealized situation Small perturbations(h/, v)about steady level flight (ho, vo) h=h+b0=v0+ L=W(mg) for v=v0, but L N v2 ≈(1+2-+..)
The Phugoid Idealized situation : • Small perturbations (h ′ , v′) about steady level flight (h0, v0) h = h0 + h ′ v = v0 + v ′ • L = W (≡ mg) for v = v0, but L ∼ v 2 , L mg = v 2 v 2 0 ≈ v 2 0 v 2 0 (1 + 2 v ′ v0 + . . .) 8
The Phugoid(cont'd) e Vertical momentum equation mh=L-mg g(1+ Energy conservation T= D mgh +muo=mgh+mu2 (to first order)- gh+uov=0 ● Equations of motion 2 +2当h 0 i+2
The Phugoid (cont’d) • Vertical momentum equation m¨h = L − mg ¨h = g(1 + 2 v ′ v0 − 1) • Energy conservation T = D mgh0 + 1 2 mv2 0 = mgh + 1 2 mv2 (to first order) → gh′ + v0v ′ = 0 • Equations of motion ¨h ′ + 2 g 2 v 2 0 h ′ = 0 ¨v ′ + 2 g 2 v 2 0 v ′ = 0 9
The Phugoid(cont'd) h' and v' satisfy a Harmonic Oscillator equa- tion Natural frequency and period V27 Light aircraft vo 1 soft/s T~20s Solution h= Asi 0 A sin U 10
The Phugoid (cont’d) h ′ and v ′ satisfy a Harmonic Oscillator Equation Natural frequency and Period ωn = √ 2g v0 τ = √ 2π v0 g Light aircraft v0 ∼ 150ft/s → τ ∼ 20s Solution h ′ = A sin √ 2g v0 t ! v ′ = − g v0 A sin √ 2g v0 t ! 10
