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Chapter 9 Oscillation 9-1 Simple Harmonic Motion, Amplitude, Period and Frequency, Phase 9-2 Rotating Vector 9-3 Single Pendulum and the Physical Pendulum 9-4 Energy in Simple Harmonic Motion 9-5 Superposition of Simple Harmonic Motions *9-6 Damped Oscillation , Forced Oscillation and Resonance 9-7 Electromagnetic Oscillation Chapter 9 Oscillation § 9-1 Simple Harmonic Motion, Amplitude, Period and Frequency, Phase Reason of Mechanical Oscillation •Restoring force •Inertia Simple Harmonic Motion 1. The relationship between displacement and time: When an object in the simple harmonic motion, its displacement is a cosine function of time. x x A cos( t ) t Simple Harmonic Motion 2. Dynamics descriptions A object goes into a back and forth motion on both side of the equilibrium position under elastic force that is directly proportional to the displacement x to the equilibrium position. F o m x x Simple Harmonic Motion 3. kinematics descriptions F kx ma k a x m The acceleration of the spring oscillation is proportional to the magnitude of the displacement and has opposite direction to the displacement. F o m x x Simple Harmonic Motion According to Newton’s second law: 2 d x F m 2 k x dt We obtain: 2 d x k x 2 m dt k Assume m 2 d x 2 x 2 dt Simple Harmonic Motion Equation of simple harmonic motion and its solution d 2x k x 2 dt m Assume: 2 k m 2 d x k x0 2 dt m 2 d x 2 x 0 2 dt The differential equation of the simple harmonic motion Its solution: where x A cos t and A are the integral constants Equation of simple harmonic motion Simple Harmonic Motion Results: (1)The vibration of the spring oscillator is simple harmonic motion. k Angular frequency: (2) m Period: T 2 2 m k (3)The angular frequency and period of the spring oscillator are only depend on the physical properties of the oscillating system. These kinds of period and frequency that only depended on the properties of the oscillating system are called natural period and natural frequency. Simple Harmonic Motion Velocity: dx v A sin( t ) v m cos( t ) dt 2 Acceleration: dv a 2 A cos( t ) am cos( t ) dt ω2 A A x, v, a a ωA O x O v t A T Physical Quantities for Simple Harmonic Motion Description x A cos t A :Amplitude,(the maximum displacement,x =±A ) The absolute value A of the maximum position of the object to its equilibrium position the amplitude. x A o A xt T T 2 t Physical Quantities for Simple Harmonic Motion Description 2 Angular Frequency : 2 T Frequency :the number of complete cycles or vibrations per unit of time. Period T:the time that the object takes to finish one complete cycle of motion. A cos t A cos t T cos t cos t T 2 T 2 , T 1 2 T 2 Physical Quantities for Simple Harmonic Motion Description Phase : ( t + ) Initial Phase : (t = 0) Phase is a physical quantity, which decides the motion state of an object in a simple harmonic motion. x A o A xt T T 2 t Physical Quantities for Simple Harmonic Motion Description Velocity Expression dx v A sin( t ) v m cos( t ) dt 2 vm A velocity is leading to displacement of /2。 Acceleration Expression dv 2 a A cos( t ) am cos( t ) dt am A 2 acceleration is out of phase with displacement. Method of Characteristics of Simple Harmonic Motion i、Mathematic Equation: x A cos(t ) ii、Diagram of Rotating Vector: x/cm iii、Diagram x-t (or v-t, a-t) 0 x 2 cos( 4π 2π t ) cm 3 3 1 2 1 t/s The solution of constant A and x A cos( t ) v A sin( t ) Initial condition: A x02 v 2 0 v0 tan x0 2 t 0 x x0 v v0 For given oscillating system, the period (or frequency) is determined by the oscillating system itself, while the amplitude and the initial phase are determined by the initial condition. Example When t 0, x 0, v 0 0, Try to resolve π v 0 A cos 2 v 0 A sin 0 sin 0 π therefore 2 π x A cos( t ) 2 x o x A o A xt T T 2 t §9-2 Rotating Vector The projection component of the vector A on Ox axis is : x A cos( t ) Conclusion: The motion of the projecting point P of the vector A can represent the simple harmonic motion along x axis. Rotating Vector • Circular frequency:the angular velocity of rotating vector • Amplitude:the absolute value of rotating vector A • Phase: ( t+ ) y • Initial Phase: (t = 0) A P t o • After one period, the phase varies 2π, everything will go on repeatedly. M Period: T 2 x Rotating Vector vm y t O an π t 2 v a A x A cos(t ) v m A v A sin( t ) x an A 2 a A 2 cos(t ) Rotating vector diagram and graph of x versus t for simple harmonic motion. Method of Resolution According to initial condition, try to resolve Amplitude and Initial phase (1)Assume time t = 0, displacement x = x0, and velocity v = v0 From Eq. x A cos ( t ) , we have xo A cos From Eq. v A sin ( t ) , we have v o A sin Method of Resolution xo A cos vo A sin v 2 2 2 2 x A (sin cos ) A 2 o 2 o 2 A x0 2 v0 vo tg xo 2 Method of Resolution Analysis: xo A cos vo A sin x 0 0, v0 0 x 0 0, v0 0 x 0 0, v0 0 x 0 0, v0 0 In the fourth quadrant In the first quadrant In the third quadrant In the second quadrant Or analysis using rotating vector diagram: After confirming x0, draw the rotating vector diagram, and then confirm the direction of velocity, v0 0, or v0 0 at last we could obtain x Example An object is in simple harmonic motion along x axis,its amplitude is 12cm,the period is 2s, at the initial moment (t = 0) the object is at the position of 6 cm and moves toward the positive direction of Ox axis. Try to resolve (1) the equation of oscillation; (2) At t = 0.5 s,Where is the object located、and what is the velocity and acceleration; (3) Assume the object was located at x = - 6 cm and moves toward the negative direction of x axis, what is the minimum time when the object moves form the position to equilibrium position. Example Solution:(1) First we need to obtain the equation of oscillation. Assume x A cos ( t ) Aaccording to this sample:A =0.12 m , T = 2 s, we have 2 s 1 T Substitute t = 0 , x0 = 0.06 m into equation of simple harmonic motion with v0 > 0, we have 0.06 =0.12 cos Example 0.06 =0.12 cos 1 cos 2 3 y 3 v 0 A sin 0 sin 0 3 x 3 Therefore, the equation of oscillation is x 0.12 cos( t 3 ) Example (2) At t = 0.5 s,we have the object located、the velocity and acceleration: x t 0 .5 0.12 cos( t ) t 0.5 0.104 m 3 dx v t 0 .5 0.12 sin( t ) t 0.5 0.189 m s 1 dt t 0.5 3 dv 2 2 a t 0 .5 0.12 cos( t ) t 0.5 1.023 m s dt t 0.5 3 Example (3) Assume at t1, x = - 0.06 m Substitute them into the motion equation: 0 .06 0 .12 cos ( t1 3 ) 2 4 t1 or 3 3 3 2 t1 t1 1 s 3 3 3 11 t2 t2 s 3 2 6 11 5 t t 2 t1 1 s 6 6 2 3 4 3 x Example Method 2:to resolve using rotating vector 3 2 5 2 3 6 t 5 5 t 6 6 2 3 3 / 2 x Example Sample: Two objects are in simple harmonic motion with same frequency and amplitude along same direction. At the initial moment the object is at the position of x1= A/2 处 and moves toward the negative direction, the other object is at the position of x 2= -A/2 and moves toward the positive direction. Try to resolve the difference of phase between the two objects. x1 A cos ( t 1 ) A 2 A cos ( t 1 ) t 1 3 Solution: v1 A sin ( t 1 ) 0 sin ( t 1 ) 0 t 1 3 Example -A -A/2 o A/2 A A 2 A cos( t 2 ) t 2 2 3 v 2 A sin ( t 2 ) 0 sin t 0 t 2 2 3 2 ( t 1 ) ( t 2 ) ( ) 3 3 Example Method 2:to resolve using rotating vector x Example Sample:An object is in simple harmonic motion as shown in the following figure, what the period is x(m) (A)2.62 s 4 2 (B)2.40 s t(s) 1 2 4 cos (D)0.382 s 3 0 4 cos 5 6 (C)0.42 s 3 2 12 T 5 2 §9-3 Single Pendulum and the Physical Pendulum 1. Single pendulum O Dynamics: d s mg sin m 2 dt 2 l T s l d mg sin ml dt 2 2 5 mg sin Single Pendulum g d 2 2 dt l g d 2 0 2 dt l Conclusion:single pendulum is a simple harmonic motion. 0 cos t g l T 2 l g Application: We can measure the period of single pendulum to obtain the gravity acceleration at this place. Physical pendulum M l F M mgl sin d 2 J J dt 2 d 2 mgl J dt 2 ω O l *C P (C is the center of quality) mgl Assume J 2 d mgl ( 5 ) 2 dt J 2 Physical pendulum d 2 2 2 dt ω O m cos( t ) l mgl J T 2π Application *C 2π J mgl P (C is the center of quality) • To calculate gravity acceleration • To calculate moment of inertia §9-4 Energy in Simple Harmonic Motion Take the spring oscillator as an example v o x x 1 1 2 Kinetic energy:E mv m 2 A 2 sin 2 ( t ) k 2 2 Potential energy:E 1 kx 2 1 kA2 cos 2 ( t ) p 2 m k 2 2 Energy in Simple Harmonic Motion x A cos ( t ) 1 E k m 2 A 2 sin 2 t 2 1 2 E p kA cos 2 t 2 x t O E m 2 k Ep The total mechanical energy: E Ek E p Ek O 1 2 1 1 2 2 2 E kA m A mv m 2 2 2 t Graph of potential energy for a simple harmonic motion Ep 1 2 Potential Energy E p kx 2 Kinetic Energy E Ek Ek E E p Ep A Simple harmonic is a motion with invariable amplitude. O 1 E kA2 2 A x Example When the displacement is 1/2A,what are the kinetic energy and potential energy? At which position kinetic energy is equal to potential energy? 1 2 Solution: E E p E k kA 2 2 1 2 1 A 1 当 x A 2 时: E p kx k E 2 2 2 4 1 2 kx0 2 3 Ek E E p E 4 1 1 1 2 A 0.707 A kA x0 2 2 2 §9-5 Superposition of Simple Harmonic Motions 1. Superposition of two simple harmonic motions with same frequency in the same direction There are two simple harmonic motions in the same direction, both of their angular frequency are ω, the amplitudes and initial phases are A1, A2, and φ1, φ2, respectively. Their motion equations are: A2 x1 A1 cos( t 1 ) x 2 A2 cos( t 2 ) 2 O x2 1 A1 x1 The phase difference 2 1 = Constant x Superposition of two simple harmonic motions with same frequency in the same direction A A1 A2 x x1 x2 x A cos( t ) A2 2 x2 A A1 1 x1 x x ( 2 1 ) Superposition of two simple harmonic motions with same frequency in the same direction A A A 2 A1 A2 cos 2 2 1 2 2 A12 A22 2 A1 A2 cos 2 1 A A12 A22 2 A1 A2 cos y A1 sin 1 A2 sin 2 tan x A1 cos 1 A2 cos 2 Conclusion: The combined oscillation is still simple harmonic motion, whose angular frequency is the same as the frequency of the component oscillation, and its amplitude of the combined oscillation is A. Superposition of two simple harmonic motions with same frequency in the same direction A A1 A2 2 A1 A2 cos( 2 1) 2 tg (1) 1 A1 sin 1 A2 sin 2 A1 cos 1 A2 cos 2 Assume : 2 1 2k obtain : (2) 2 A k 0,1,2, A A 2 A1 A2 A1 A2 2 1 2 2 Assume : 2 1 (2k 1) obtain : A k 0,1,2, A12 A21 2 A1 A2 A1 A2 Superposition of two simple harmonic motions with same frequency in the same direction (1) Phase difference 2 1 2 k x oA 1 A2 k 0, 1, 2, x o T t A x ( A1 A2 ) cos(t ) A A1 A2 2 1 2k π Superposition of two simple harmonic motions with same frequency in the same direction (2)Phase difference 2 1 ( 2 k 1) x x o o k 0,1,2, A1 2 T t A A2 x ( A2 A1 ) cos(t ) A A1 A2 2 1 (2k 1)π Superposition of two simple harmonic motions with same frequency in the same direction Conclusion (1)Phase difference 2 1 2 k π ( k 0 , 1,) A A1 A2 Enhance ) (2)Phase difference 2 1 ( 2 k 1) π ( k 0 , 1, A A1 A2 Weaken (3)Phase difference can be arbitrary value A1 A2 A A1 A2 Example There are two objects in simple harmonic motions in the same direction as shown in the following figure: try to resolve: 1、the amplitude of the combined oscillation 2 、the motion equations of superposition x x1 (t ) A 1 2 Resolution T A A2 A1 A2 x 2 (t ) T A2 A1 cos 1 0 1 1 2 2 A A2 cos 2 0 2 2 A1 2 2 2 x A2 A1 cos( t ) 2 T 2 t Example A A A2 sin sin 6 6 A 20 sin sin sin 1 A2 6 10 6 2 A2 A1 2 1 2 Superposition of two simple harmonic motions with same frequency in perpendicular direction Simple harmonic motion along x direction x A1 cos( t 1 ) y A 2 cos( t 2 ) Simple harmonic motion along y direction Superposition of two simple harmonic motions with same frequency in the same direction x A1 cos( t 1 ) y A2 cos( t 2 ) y x cos t cos 1 sin t sin 1 A1 y cos t cos 2 sin t sin 2 A2 2 x 2 x y 2 xy 2 2 cos( 2 1 ) sin ( 2 1 ) 2 A1 A2 A1 A2 Conclusion:the superposition of two simple harmonic motions in perpendicular directions with the same frequency is ellipse motion, which is determined by phase difference. Superposition of two simple harmonic motions with same frequency in the same direction Discussion: (a) 2 1 0 (or 2k ) 2 2 x y 2 xy 2 0 2 A1 A2 A1 A2 y 2 x y 0 A1 A2 A2 A2 y x Slop 0 A1 A1 Conclusion:the trajectory of the combined oscillation is a straight line. x Superposition of two simple harmonic motions with same frequency in the same direction x 2 y 2 2 xy 2 cos( ) sin ( 2 1 ) 2 1 2 2 A1 A2 A1 A2 (b)当:2 1 2 2 or 2 k 1 2 2 x y 2 1 2 A1 A2 Conclusion:the trajectory of the combined oscillation is a perfect ellipse. y x Superposition of two simple harmonic motions with same frequency in the same direction 2 2 x y 2 xy 2 2 cos( 2 1 ) sin ( 2 1 ) 2 A1 A2 A1 A2 (c) 2 1 2 k 1 2 y 2 x y 2 xy 2 0 2 A1 A2 A1 A2 2 x y 0 A1 A2 x A2 A2 y x , Slop : 0 A1 A1 The trajectory of the combined oscillation is a straight line. Superposition of two simple harmonic motions with same frequency in the same direction The superposition of two simple harmonic motions in perpendicular directions with the same frequency but different phase difference. Superposition of two simple harmonic motions with same frequency in the same direction Lissajou figure Superposition of several simple harmonic motions with same frequency in the same direction x1 A1 cos t 1 x2 A2 cos t 2 x3 A3 cos t 3 A3 A A2 A1 A A A , tan 2 y 2 1 x A cos t 2 x 3 Ay Ax A x A1 cos 1 A2 cos 2 A3 cos 3 A y A1 sin 1 A2 sin 2 A3 sin 3 Superposition of several simple harmonic motions with same frequency in the same direction x1 A0 cos t A x 2 A0 cos( t ) o A1 A2 A3 A A5 x x3 A0 cos( t 2 ) A Ai NA0 i x N A0 cos[ t ( N 1) ] 4 (1) 2k π Discussion ( k 0, 1, 2, ) (2) N 2k ' π ( k ' kN , k ' 1, 2 , ) A4 A3 A 2 O A6 A x 1 A5 A0 Superposition of two simple harmonic motions with different frequency in the same direction beat The rotating angular speed of A1 with respect to A2 is 2 1 1 A A1 A2 A2 O 2 A 1 A1 x 2 For the superposition of two simple harmonic motions in the same direction when the frequencies of two component oscillations are relatively large while difference of the two frequencies is very small, the phenomenon of the resultant amplitude ever and agah strengthening and weakening is called the beat. Superposition of two simple harmonic motions with different frequency in the same direction beat Superposition of two simple harmonic motions with different frequency in the same direction beat 2 The period of the beat:T 2 1 2 1 2 1 Beat frequency: 2 Assume : A1 A2 A x1 A cos(1t ) x 2 A cos( 2 t ) x x1 x 2 A cos(1t ) A cos( 2 t ) 2 1 2 1 2 A cos t cos( t ) 2 2 Superposition of two simple harmonic motions with different frequency in the same direction beat When 2 1 2 1 2 1 2 1 t cos t We obtain: x 2 A cos 2 2 Superposition of two simple harmonic motions with different frequency in the same direction beat 2 1 2 1 x 2 A cos t cos t 2 2 Assume : 0 2 1 2 1 x 2 A cos 2 t cos 2 t 2 Amplitude:2 A cos 2 2 1 2 2 t The amplitude of the combined oscillation various slowly and periodically with the time. 2 1 cos 2 t 2 Superposition of two simple harmonic motions with different frequency in the same direction beat The second method:Superposition of rotating vector ( 2 1 )t ( 2 1 ) A 2 2 2t 2 1t 1 o 1 2 0 x2 A 2 1 1 A1 x1 x 2 π ( 2 1 )t x Superposition of two simple harmonic motions with different frequency in the same direction beat Amplitude A A1 2(1 cos ) 2 1 2 A1 cos( Beat frequency Angular frequency x1 x2 cos t A 2 2 t) 2 1 ( 2 1 )t A2 o A1 1 x2 1t 2t 1 2 2 A x x 2 1 x1 Superposition of two simple harmonic motions with different frequency in the same direction beat Applications: • Acoustics • Speed measurement; • Radio-technology; • Satellite tracking §9-7 Electromagnetic Oscillation The phenomenon that charges and current, electric field energy and magnetic field energy vary with time periodically is called Electromagnetic Oscillation. L L + C E ε C I0 Q0 Q0 L S LC Circuit I0 L C B + Q0 C A B C E Q0 L B C D 1. The free electromagnetic oscillation without damping 2. The equation of the free electromagnetic oscillation without damping The self induction electromotive force is : dI q L dt C q VC C I L dI q Equation of Circuit: L dt C 2 d q 1 dq I q 2 dt dt LC C 2. The equation of the free electromagnetic oscillation without damping Charge: q q 0 cos t 1 LC 1 Frequency: Period:T 2 LC 2 2 LC dq q 0 sin t Current: I dt I I max sin t I max q 0 q0 Voltage: V cos t C Conclusion:In the circuit of LC,the current、Voltage、 charges are in simple harmonic motions. The variation of charges and current with time for a free electromagnetic oscillation without damping q i Q0 I 0 O π 2 ﹡ π ﹡ 2π (t ) π q Q0 cos( t ) i I 0 cos(t ) 2 3. The energy of the free electromagnetic oscillation without damping 2 q q Electric field energy: W e 0 cos 2 t 2C 2 C 2 2 2 L q 1 2 0 Magnetic field energy:W m LI sin 2 t 2 2 q 02 2 sin t 2C The total energy of LC oscillation circuit: 2 0 q W We W m 2 C In the process of free electromagnetic oscillating without damping, the electric field energy and the magnetic field energy converts into each other constantly, while at any moment the sum of them remains unchanged. 3. The energy of the free electromagnetic oscillation without damping The conditions of conservation of electric and magnetic field energy in LC oscillation circuit: 1. the resistances in the circuit must be zero ; 2. there are not any electromotive forces existing in the circuit; 3. the electromagnetic energy can not be radiated in the form of electromagnetic wave. Therefore, LC electromagnetic oscillation circuit is an ideal model of oscillation circuit. LC electromagnetic oscillation compared with simple oscillation motion Equation of LC electromagnetic oscillation Equation of simple oscillation motion d 2q 2 q 2 dt d2x 2 x 2 dt q Q0 cos( t ) dq i Q0 sin( t ) dt x A cos( t ) dx v A sin( t ) dt Example In a LC circuit, it is known L = 260 μH,C = 120 pF, at the beginning the electric potential difference between the two plates of the condenser is U0 = 1V and the current is i0 = 0. Try to resolve Solution(1)Oscillation frequency: (2)The maximum current 当t = 0 时 1 2 π LC 9 .01 10 5 Hz q0 Q0 cos CU 0 i0 Q0 sin 0 C I 0 Q0 CU 0 U 0 0.679 mA L Example (3)the relation that the electric field energy in between the two plates of the condenser varies with time: 1 2 2 10 2 E e CU 0 cos t (0.60 10 J ) cos t 2 (4)the relation that the magnetic field energy in the self induction coil varies with time: 1 2 2 10 2 E m LI 0 sin t (0.60 10 J ) sin t 2 1 10 (5) E e E m 0.60 10 J E e 0 CU 02 2 Therefore at any time the sum of the electric field energy and the magnetic field energy equals the electric energy at the initial time.