Oscillations. Tacoma Narrow Bridge: Example of Torsional Oscillation

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1 Oscillations Mechanical Mass-spring system nd order differential eq. Energy tossing between mass (kinetic energy) and spring (potential energy) Effect of friction, critical damping (shock absorber) Simple pendulum, physical pendulum (sweet points) Tacoma Narrow Bridge: Example of Torsional Oscillation

2 Mass-Spring System Restoring force k dv d dx d x ( t) ( ) d x t by the spring: : spring force constant in N/m Try x k + x ( t ) = dt M = Acos ωt. (N) Equation of motion for the mass: mass acceleration = F = kx M = M = M = kx dt dt dt dt d d Acosωt = Aωsin ωt, Acosωt = Aω cosωt dt dt Subsitution to the diff. eq. yields MA cos t kacos t ω ω + ω = ω = ω = k M k M (rad/sec) oscillation angular frequency F

3 Example: A spring m longhanging from the ceiling elongates by 3 cm when a mass of.5 kg is attached to the end. If the mass is pulled down another 5 cm and released, what is the oscillation frequency? The spring constant is from F Mg N F = k l k = = = = 49 N/m l l.3 m k 49 N/m ω= = = 5.7 rad/s, f =.9 Hz M.5 kg Mg g 9.8 m/s Alternativley, ω = = = = 5.7 rad/s M l l.3 m Example: Discuss energy exchange in the above Example. Initial potential energy is k ( x) = 49 (.5) =.6 J. At time t, kx cos ωt Kinetic energy of the mass is dx M = Mω x sin ωt = kx sin ωt dt The total enengy: kx ( cos ωt + sin ωt) = k x =const.

4 ( ) Energy Tossing Potential energy (red) + kinetic energy (green) = constant (black) If x t = x cos ωt, dx + = cos + sin kx M kx ωt Mω x ωt dt = kx ( cos ωt + sin ωt) = kx x

5 Energy conservation oscillation equation Mv + kx = const. d dv dx Mv + kx = Mv + kx = dt dt dt dx dv d x v =, = dt dt dt d x d x k dt dt M Then M + kx = + x = Effect of Friction d x dx M + f + kx = damping (dissipation of energy) dt dt Compare with LCR circuit d q R dq q dt L dt LC = + +

6 Damped oscillation f k x'' + x' + x = M M x" + γx' + ωωx = ( a) γ =.ω ( b) γ =.5ω f k γ =, ω = M M From top ( c) γ = ω x(t) x(t) t t Solution γ t γ x( t) = xe cosωt+ sin ωt ω ω = ω γ x(t) t

7 Common force d M ( x + x) = kx dt d F= kx = kx k M + x= kx dt k d kk M x x k x Two springs = = eff, = + dt k+ k keff k k In this case, the force is additive, eff ( ) F= kx kx = k + k x k = k + k

8 Pendulum Pendulm of length L and mass M Torque about the pivot MgLsin θ MgLθ (small θ) I d θ d θ g MgLθ dt dt L + = + θ = Oscilaltion frequency ω = g L independent of mass Example: If L = m, g ω = = 9.8 rad/s =3.3 rad/s L f =.498 Hz The oscillation frequency can be used for measuring g.

9 Nonlinear Oscillation The pendulum equation d θ g + sinθ = dt L is nonlinear. The sine function can be expanded for small θ as 3 sinθ θ θ + = θ θ 6 6 d θ dt g L 6 Then + θ θ = and the oscillation frequency is modified a g ωθ ( ) = θ /6 L where θ is the amplitude which reduces the oscillation frequency. g More advanced calsulation yields ωθ ( ) = θ / 8 L s

10 Numerical solutions small and large amplitude theta t y x

11 Oscillation involving moment of inertia A cylinder attached to a spring rolls on floor. Energy conservation is Mv + Iω + kx = const. where I = Ma 3 = ω + = 4 and v a. Then Mv kx const. Oscillation frequency is ω= 3 k M (moment of inertia of cylinder)

12 A stick of length L = 3 LMg sinθ LMgθ d θ I = LMgθ dt d θ I + LMgθ = dt Stick pivoted at its end is pivoted at its end. The moment of inertia about the end is I ML. The restoring torque for small is LMg 3g ω= = I L If L = m, ω= 3.83 rad/sec, f =.6 Hz θ θ -Mgθ Mg L

13 Circular loop The moment of inertial of a circular loop about a point on itself is I = I + Ma = Ma + Ma =Ma CM Then the oascilaltion frequency g ω= a If fact, an incomplete loop also has the same oscillation frequency. (Homework) pivot θ CM Mg a θ a

14 Half Circle The moment of inertia about the pivot M π / θ I = asin adθ π a Ma π / θ = 8 sin π Ma π / Ma π = 4 ( cosθ) dθ 4 π = π CM is at a from the pivot. Then π ( / π ) ag g ω = = Ma π a 4 π dθ asinθ/ θ/ a θ/ a(cosθ cosθ ) θ θ

15 Sloshing Oscillation of Water in Pan, Lake, Bay (Seiche) The center of mass follows the trajectory H y = 6 x a where H is the depth of water and a is the channel length. The oscillation frequency is ω= H g a

16 Sloshing Oscillation (cont.) ( ) Let the water displacement at the edge be h t. The CM of the rectangle BDEC is at x =, y = ( H h) The CM of the triangle ABC is at a x =, y = h+ H h = H h Then the CM of the entire water is at a a a H h x = ah = h, y = ( H h) ah H h ah 6 6 H ah + 3 = + 6 H H Eliminating h, we find y = 6 x a ω = Hg a A ball rolling along a parabolic curve y = Ax oscillates at ω = Ag y = Ax

17 Sweet Point Sweet point of baseball bat, tennis racket, etc. can be determined as follows. The grip point can be regarded as a pivot for physical pendulum. Its oscillation frequency is ω= I CM Mgd + Md g Simple pendulum of length L : ω'=. From ω = ω', L ICM + Md ICM L = = + d Md Md This determines the position of sweet point. Force acting at the grip (pivot) becomes minimum if the ball is hit at the sweet point. (Homework)

18 Electromagnetic Oscillations Charged capacitor is suddenly connected to an inductor. Voltage balance is + L = But i = + q= q di dq d q C dt dt dt LC The charge oscillates at ω = LC If a resistance is inlcuded, L Ri q = q di d q R dq + + = + + C dt dt L dt LC Weakly damped solution is γ t R - q( t) = qe cos ωt, γ = (s ) L

19 The capacitor stores energy of CV = The inductor stores energy of Li q C q For cos, cos C C q q( t) = q ωt = t ( ) dq t q = = sin = sin Li L Lω q ωt ωt dt C Total energy is constant as in spring-mass mechanical oscillation. The capacitor and inductor exchange energy periodically. ω

20 Example: 5 µ F capacitor charged to kv is suddenly connected to a µ H inductor. Assuming a 5 m Ω circuit resistance, describe how the initial energy decays. 3 R 5 = = 6 L 5 - ω = =3.6 s. Weakly damped. LC s is smaller t γ t 3 5t 5 ( ) = cosω = cos ( 3.6 ) V t Ve t e t Intial energy CV =.5 J. Then 5t 5 ( ) =.5 cos ( 3.6 ) E t e t han.5 y.5 5e t

21 Forced Oscillation If an LCR circuit is driven by an oscillator with voltage V cos ωt, the amplitude of the current is ( ω) V i = which exhibits a resonance ωl + R ωc when ω =. LC Example: C = mf, L = mh, R =. Ω V i = V = 3 ω. 3 + ω 8 6 y x

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