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Chapter 9 Oscillation 9-0 Basic Requirements of the Chapter Teaching 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 Brief Summary of this Chapter Basic Requirements of the Chapter Teaching 1、To master physical significance of physical quantities that describe simple harmonic motion and the relations between them. 2 、 To master the representation of rotating vector for simple harmonic motion, and be able to analysis and resolve problems of simple harmonic motion. 3、To master the basic characteristics of simple harmonic motion, be able to establish the differential equation of simple harmonic motion, to obtain the motion function of the simple harmonic motion determined by the initial condition, and to understand its physical significance. Basic Requirements of the Chapter Teaching 4、to understand the energy converting process of simple harmonic motion, and be able to calculate the mechanical energy of a simple harmonic motion. 5、To understand the regulation of superposition of simple harmonic motions with same frequency in the same direction, to grasp the characteristics of beat and superposition of two simple harmonic motions with the same frequency in perpendicular direction. 6、To grasp the physical principle of electromagnetic oscillation. § 9-1 Simple Harmonic Motion, Amplitude, Period and Frequency, Phase Oscillation is a periodical motion. If a periodical motion is only the back-and-forth motion of the position, which is called Mechanical Oscillation. Examples for Oscillation: 1. The back-and forth motion of objects (ex. The spring oscillation and single pendulum) 2. The cycle of current and voltage 3. Atoms vibration within a molecule. § 9-1 Simple Harmonic Motion, Amplitude, Period and Frequency, Phase Reason of Mechanical Oscillation •Restoring force •Inertia The Characteristics of Simple Harmonic Motion 1. The relation 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 The Characteristics of Simple Harmonic Motion 2. Kinematics 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 The Characteristics of Simple Harmonic Motion 3. Dynamics 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 Every complicated oscillation is the superposition of several simple harmonic motions. Simple Harmonic Motions Superposition Decomposition Complicated Oscillation The Regulation of Simple Harmonic Motion (Take the spring oscillator as an example) The oscillation system including the spring and the object is called the spring oscillation. F o x x According to Hooke’s law: F k x (where k is spring constant) (1) The elastic force that the object subjected is directly proportional to the displacement x to the equilibrium position. (2) The direction of the elastic force is opposite to the direction of the displacement, and orients to the equilibrium position. The Regulation of 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 The Regulation of 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 The Regulation of Simple Harmonic Motion Results: (1)The vibration of the spring oscillator is a simple harmonic motion. k (2)Angular frequency: 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. The Regulation of 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 is 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. Methods of Characteristics for Simple Harmonic Motion i、Mathematic Equation: x A cos(t ) ii、Diagram of Rotating Vector: 1. To determine the phase and initial phase of a simple harmonic motion by rotating vector. 2. To discuss the phase difference of two simple harmonic motions by rotating vector. iii、Diagram x-t (or v-t, a-t) x 2 cos( 4π 2π t ) cm 3 3 x/cm 0 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 Rotating vector diagram and graph of x versus t for a simple harmonic motion. The graph of x versus t for simple harmonic motion (1) The graph represents the regulation of the simple harmonic motion. The direction of velocity is shown in the graph. (2) The graph reflects period T, amplitude A, initial phase and phase. The graph of x versus t for simple harmonic motion (3) The graph shows the relation between displacement and time: x A cos t x t (4) The graph shows the relation between phase difference and time: t x The graph of x versus t for simple harmonic motion x (4) The graph indicates the direction of the force that the object subjected and the direction of the acceleration. F = - kx,The direction of the force is opposite to the direction of the displacement, and the acceleration has same direction to the force. (5) The graph indicates the position that object has maximum and minimum kinetic energy and potential energy. When the displacement is zero, kinetic energy is the maximum, while potential energy is zero. When the displacement reaches maximum, potential energy is the maximum, while kinetic energy is zero. Methods 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 Methods 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 Methods 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, we draw the rotating vector diagram, and then confirm the direction of velocity, v0 0, or v0 0 at last we could obtain x Methods of Resolution (2)Assume time t = 0,xo = kA, where k is a constant, given the direction of velocity, we obtain: x A cos t k cos kA A cos has two values that can be determined by the direction of velocity from rotating vector diagram. If given initial condition, we can obtain the integral constants A and by equation (displacement or velocity). (3)If given xt, vt at time t, we can obtain the integral constants A and by equation (displacement , velocity or acceleration), especially using the following expressions: v max A a max A 2 Examples An object is in simple harmonic motion along x axis,its amplitude is 12 cm,the period is 2 s, 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 from the position to equilibrium position. Examples Solution:(1) First we need to obtain the equation of oscillation. Assume x A cos ( t ) According 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 Examples 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 ) Examples (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 Examples (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 Examples Method 2:to resolve the problem using rotating vector 3 2 5 2 3 6 t 5 5 t 6 6 2 3 3 / 2 x Examples Sample: Two objects are in simple harmonic motion with same frequency and amplitude along the same direction. At the initial moment the object is at the position of x1= A/2 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 phase difference 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 Examples -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 Examples Method 2:to resolve the problem using rotating vector x Examples Sample: An object is in simple harmonic motion along x axis,at the initial moment (t = 0) the object is at the position of 0.04 m along the positive direction of Ox axis. Try to resolve (1) the equation of oscillation; (2) The velocity where the object was located at x = A/2 at the first time when the object moves from the initial position to the position. Examples Solution: (1) According to this sample, x 0 0.04m , v0 0 , 6 .0 s Substitute them into the following expression, we have Amplitude: A x0 2 v0 2 0 2 x0 0 . 04 m v0 0 Initial phase arctg x0 then: x 0.04 cos (6.0t ) m Examples (2) First we need to obtain phase x A cos( t ) x t arccos A A2 1 5 t arccos arccos (or ) A 2 3 3 A x A x , t 2 3 Therefore, v A sin t 0.04 6.0(sin ) 3 1 0.208 m s Examples Method 2:to resolve the problem using rotating vector Solution: According to this sample t 3 A x A x 2 v A sin t 0.04 6.0(sin 0.208 m s 1 3 ) Examples 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 mass) 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 *C 2π Applications J mgl P (C is the center of mass) • 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 E m 2 k Ep The total mechanical energy: E Ek E p t O Ek O t 1 2 1 1 2 The restoring force is 2 2 E kA m A mv m a conservative forces 2 2 2 Summary (1) Both the kinetic energy and potential energy of the system vary in period with time, while the total energy remains unchanged. When the displacement is zero, kinetic energy is the maximum, while potential energy is zero. When the displacement reaches maximum, potential energy is the maximum, while kinetic energy is zero. (2) The frequency of kinetic energy and potential energy variation is two times of the frequency of vibration of the spring oscillator. (3) The total mechanical energy of a simple harmonic oscillator is proportional to the square of the amplitude. 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 motion 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 At x A 2:E p kx k E 2 2 2 4 3 Ek E E p E 4 1 1 1 1 2 2 A 0.707 A kx0 kA x0 2 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 a 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) If the phase difference is x oA 1 A2 2 1 2 k 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) If the phase difference is 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 Summary (1)If the phase difference is 2 1 2 k π ( k 0 , 1,) then A A1 A2 Enhance (2)If the phase difference is 2 1 ( 2 k 1) π ( k 0 , 1,) A A1 A2 Weaken (3)The 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 equation 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 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 perpendicular 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 perpendicular direction Discussion: (a)If 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 perpendicular direction x 2 y 2 2 xy 2 cos( ) sin ( 2 1 ) 2 1 2 2 A1 A2 A1 A2 (b)If: 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 perpendicular direction 2 2 x y 2 xy 2 2 cos( 2 1 ) sin ( 2 1 ) 2 A1 A2 A1 A2 (c)If 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 perpendicular 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 perpendicular 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 The period of the beat: Beat frequency: Assume : 2 T 2 1 2 1 2 1 2 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 If 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 is a harmonic oscillator. 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 The free electromagnetic oscillation without damping 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 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 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: q 02 W We W m 2C 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. 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 harmonic motion Equation of LC electromagnetic oscillation d 2q 2 q 2 dt q Q0 cos( t ) dq i Q0 sin( t ) dt Equation of simple harmonic motion d2x 2 x 2 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 At t = 0 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 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. Brief Summary of this Chapter 1、 Characteristics of simple harmonic motion (1) Definition of simple harmonic motion (2) Kinematics and dynamics descriptions (3) The velocity and acceleration of the oscillator (4) The period, frequency and angular frequency of the oscillator (5) The relation between amplitude, initial phase and initial displacement and initial velocity (6) The characteristics of mechanical energy Brief Summary of this Chapter 2、Rotating vector (1) to determine the phase and initial phase for simple harmonic motion using rotating vector. (2) to discuss the phase difference of two simple harmonic motions. 3、The superposition of simple harmonic motions (1) to calculate the amplitude and phase difference of the combined oscillation, the component oscillations have same frequency and same direction. (2) Superposition of two simple harmonic motions with same frequency in perpendicular direction,the corresponding chart in the special phase difference. (3) The engender of beat and the calculation of beat frequency. Brief Summary of this Chapter 4、 Electromagnetic Oscillation (1) to understand the physical diagram of electric field and magnetic field oscillation in LC oscillation circuit. (2) to understand the differential equation of LC oscillation circuit and to master the formula of oscillation frequency for LC oscillation circuit.