Oscillation - Tran Thi Ngoc Dung

Nội dung text: Oscillation - Tran Thi Ngoc Dung

1. OSCILLATION Tran Thi Ngoc Dung – Huynh Quang Linh – Physics A2 HCMUT 2016
2. Contents • Simple Harmonic Motion • The Physical Pendulum • Damped Oscillations • Driven Oscillations and Resonance • Using Complex Numbers to Solve the Oscillator Equations
3. 1.1 Simple harmonic motion Spring pendulum O Newton’s 2nd law: x ma P N fspring (1) f d2x s (1)/x : m 2 kx dt x 2 d x 2 x ox 0 o k / m dt 2 x : distance from A: amplitude. equilibrium x Acos(ot ) (ω t +  ): phase of the motion v A sin( t ) Velocity phase constant. o o The phase constant depends on the 2 Acceleration: a Ao cos(ot ) choice at t = 0. 1 2 1 2 2 Kinetic Energy KE mv kA sin (o t ) 2 2 1 1 Potential Energy PE kx 2 kA 2 cos2 ( t ) 2 2 o The total energy in simple harmonic 1 motion is proportional to the square of W KE PE kA 2 const the amplitude. 2 Average Kinetic Energy =Average 1 2 Potential Energy KE PE kA 4
4. The Physical Pendulum O A rigid object pivoted about a point other than its center of mass will oscillate when displaced from D equilibrium. Such a system is called a physical  pendulum. CM I  Dsin d2θ mg IΔ mg Dsinθ dt 2 2 d θ mg D mg D mg D sinθ 0 ωo ; sinθ θ : small angle ω dt 2 I I o Δ Δ IΔ 2 d θ 2 ω  0 2 I 2 o Period of a Phys. T 2 dt pendulum o mg D  o cos(ot ) Uniform disk Simple pendulum 1 3 2 2 2 2 I m I mR mR mR G 2 2  3  mR 2 m T 2 I 3R T 2 2 2 2 g mg D mg R 2g
5. Damped Oscillations O Let to itself, a spring or a pendulum eventually stops oscillating because the mechanical energy is dissipated by x f frictional forces. The oscillation is damped drag v ma P N fspr fdrag (1) x fspring d2x x m kx - bv dt 2 d2x b dx k b x 0 ωo k/m; 2 dt 2 m dt m m 2 d x dx 2 2 ox 0 dt 2 dt t 1 m 2 2 x Ae cos(t ) τ : decay time T 2β b  2 2 2 o  2 2  β ωo :overdamped  o  o 1 T>To=2 /o o β ωo :critically damped When  is greater than or equal to o the system does not oscillate. If  = o the system is said to be critically damped; it returns to equilibrium with no oscillation in the shortest time possible. When  >o the system is overdamped.
6. t x Damped x Ae cos(t ) b ωo k/m; 2 Oscillation 2 m   2 2  1 o o  1 m o τ : decay time 2β b β ωo :overdamped β ωo :critically damped t β 0.7 ωo 10 rad/s 2 2 ω ωo β 9.9755rad/ s x e 0.7t cos(9.9755t) T=0.63s
7. Driven Oscillations and Resonance O To keep a damped system going, energy must be put into the system. When this is done, the x oscillator is said to be driven or forced. If you put fdrag energy into the system faster than it is dissipated, v the energy increases with time, and the amplitude x f F increases. If you put energy in at the same rate it is x spring driven being dissipated, the amplitude remains constant over time. Fdrivent=Focost We will discuss the general solution of ma P N fspring fdrag Fdriven (1) Equation * qualitatively. 2 It consists of two parts, the transient d x solution and the steady-state solution. m kx - bv Focost dt 2 The transient part of the solution is identical to that for a damped oscillator d2x b dx k F x o cost Over time, this part of the solution dt 2 m dt m m becomes negligible because of the exponential decrease of the amplitude. We b Fo ωo k/m; 2 ; fo are then left with the steady− state m m solution. 2 d x dx 2 x Ce t cos(t ) 2 ox focost trans dt 2 dt xsteady Acos(t ) x Acos(t )
8. Using Complex Numbers to Solve the Oscillator Equations x Acos(t ) fo d2x dx A 2 2x f cost 2 2 2 2 2 dt2 dt o o (o  ) 4  jt 2 x Ae x Rex tan  2 2 2 2 (  j2 o )x focost o jt The steady-state solution does not foe x 2 2 depend on the initial conditions o  j2 2 2 2 2 2 2 2 j Find resonance frequency and res. o  j2 (o  ) 4  e amplitude 2 tan  2 2  o 2 2 2 2 2 w ( o ) 4  f e jt dw x o 2(2 2)(2) 82 0 2 2 2 2 2 j o (o  ) 4  e d 4(2 2 22) 0 f e j(t ) o o 2 2 Resonance 2 2 2 2 2 resonance o 2 (o  ) 4  frequency fo f Aresonance x o cos(t ) 2 2 Resonance 2 2 2 2 2 2 o  (o  ) 4  Amplitude
9. Resonance curves x Acos(t ) f =0.05o A o 2 2 2 2 2 (o  ) 4  2 tan  2 2  o 2 2  0 resonance o 2 =0.25o f A o fo 2 Aresonance o 2 2 2 o  
10. Example 14-10 A uniform stick of mass m and length L is pivoted at one end. Find the period of oscillation for small angular displacements. 1 L,m mL2 I 2L T 2 2 3 2 mgD mgL / 2 3g L 1m,T 1,64s L T' 2 2s g 1. Let U be the potential energy (with the zero at zero displacement) and K be the kinetic energy of a simple harmonic oscillator. Uavg and Kavg are the average values over a cycle. Then: A. Kavg > Uavg B. Kavg < Uavg C. Kavg = Uavg D. K = 0 when U = 0 E. K + U = 0 ANS C
11. 2. A particle is in simple harmonic motion along the x axis. The amplitude of the motion is xm. At one point in its motion its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = xm, the kinetic and potential energies are: A. K = 5 J and U = 3J B. K = 5 J and U = −3J C. K = 8 J and U = 0 ans: D D. K = 0 and U = 8J E. K = 0 and U = −8J 3. Two uniform spheres are pivoted on horizontal axes that are tangent to their surfaces. The one with the longer period of oscillation is the one with: A. the larger mass B. the smaller mass C. the larger rotational inertia D. the smaller rotational inertia E. the larger radius ans: E
12. 3. Five hoops are each pivoted at a point on the rim and allowed to swing as physical pendulums. The masses and radii are hoop 1: M = 150 g and R = 50cm hoop 2: M = 200 g and R = 40cm hoop 3: M = 250 g and R = 30cm hoop 4: M = 300 g and R = 20cm hoop 5: M = 350 g and R = 10cm Order the hoops according to the periods of their motions, smallest to largest. A. 1, 2, 3, 4, 5 B. 5, 4, 3, 2, 1 C. 1, 2, 3, 5, 4 D. 1, 2, 5, 4, 3 E. 5, 4, 1, 2, 3 ans: B