收藏 分享(赏)

第10章 机械振动的电磁振荡.ppt

上传人:hwpkd79526 文档编号:10068472 上传时间:2019-10-04 格式:PPT 页数:47 大小:1.95MB
下载 相关 举报
第10章  机械振动的电磁振荡.ppt_第1页
第1页 / 共47页
第10章  机械振动的电磁振荡.ppt_第2页
第2页 / 共47页
第10章  机械振动的电磁振荡.ppt_第3页
第3页 / 共47页
第10章  机械振动的电磁振荡.ppt_第4页
第4页 / 共47页
第10章  机械振动的电磁振荡.ppt_第5页
第5页 / 共47页
点击查看更多>>
资源描述

1、1,Mechanical Oscillation AND Electromagnetic Oscillation,CHAPTER 10,By Dr. G Y Chen,第十章 机械振动和电磁振荡,3,Many oscillations exist in nature.,Oscillation of springs,Play on a swing,4,10-1 Simple Harmonic Motion (SHM),简谐振动,Two conditions of mechanical oscillation:,惯性,Free oscillations:,If the spring k is ma

2、ssless,Simple Harmonic Oscillator,Simple Harmonic Motion(SHM),简谐振子,弹性回复力,Only by the elastic restoring force,5,10-1 Simple Harmonic Motion(SHM),1 characteristics,The system (black spring) in Fig. 10-1 is called as a simple harmonic oscillator.,Spring(k): massless; body(m);,Process of SHM,6,Process o

3、f SHM,Fig. 10-2,7,Equilibrium position: point O(F=0);,物体:在平衡位置附近作周期性往复运动;,Spring: deformation(形变)-x;,Body: displacement(位移)-x;force: F=-kx (frictionless);,坐标原点:通常取为原长位置O;,According to Newtons Second Law,Some concepts,8,acceleration:,whose solution is,9,作谐振动的物体的加速度,总是与其离开平衡位置的位移大小成正比,且两者方向相反。 运动学特征。,

4、作谐振动的物体所受合力大小,总是与其离开平衡位置的位移大小成正比,且两者方向相反。 动力学特征。,10,Notes:,(1) Maximum values:,(2) The curves of x(t)、v(t) and a(t):,11,2 Three characteristic quantities(特征量) of SHM.,We can see that:,(1) The displacement is determined by three quantities: A, , ;,(2) x=Acos(t+) is a periodic function of time t.,12,(

5、1) Amplitude (A),A-the maximum displacement from the equilibrium position.,(2) The period (T) and frequency (),Period(周期) T: The time necessary for one complete oscillation(a complete repetition of the motion).,完成一次全振动所需的时间,SI Unit: s,13,Frequency(频率): The number of complete oscillation finished by

6、the particle or the system per unit time.单位时间内粒子或系统完成全振动的次数。,Apparently:,SI Unit: Hz,1 Hz=1 s-1,Angular frequency(角频率) : The number of complete oscillation finished by the particle or the system in 2 seconds.,Have a inverse relationship with T,SI Unit: rad/s,14,(3) Phase(相位),If A, and are known, the

7、 motion of the oscillator is determined completely by the quantity =t+, which is called phase.当A、已知时,谐振动的物体在任意时刻的运动状态由t+决定。t+称为相位。 When t =0, The phase turns into , this is called initial phase(初相). 而是t =0时的相位,称为初相位,简称初相。反映了振动的初始状态。,15,Difference of phases (相位差) are:=(t+2)-(t+1)=2-1 equal to the dif

8、ference of initial phase,(4) Comparison of Phases(相位的比较),Condition: the frequency should be the same.两个频率相同的谐振动才能比较步调。,If x1=A1cos(t+1 ) x2=A2cos(t+2 ),Note:,16,(5) Discussions:,(1)如果0,21,则称第二个谐振动超前于第一个谐振动的相位。 (2)如果=2-1=0或2的整数倍,则两个谐振动同时到达正的最大位移、最小位移。任意时刻,振动方向相同,称两个谐振动同相或同步。 (3)如果 =2-1=的奇数倍,则一个物体到达正的

9、最大位移,另一个物体正好到达负的最大位移。任意时刻,振动方向相反,称两个谐振动为反相。,17,3 Determination of A and .,Initial conditions:,Conclusions:,If,18,Examples(1):,一个质量为10g的物体作简谐振动,周期为4s, t=0时坐标为24cm速度为零。计算:(1)t=0.5s时物体的位置;(2) t=0.5s时物体受到的力的大小和方向;(3)从初始位置运动到x=-12cm所需的最少时间;(4)x=12cm时物体速度的大小。,解:本题已知物体作简谐振动。周期为T=4s,振幅为A=24cm,则:,19,由初始条件:,(

10、1)当t=0.5s时:,20,(2)当t=0.5s时,物体受力:,与x轴正向相反。,(3)从初始位置运动到x=-12cm的最少时间:,21,(4)x=-12cm物体的速度大小:,注意:单位换算。,22,Example(2):,In figure, prove the simple pendulum (单摆)is a simple harmonic oscillator when is small.,Prove:,23,Prove:,(1) Displacement: or x,(2) Restoring torque(恢复力矩):,(3) Angular(角量) acceleration:,N

11、ote:,(linear线量),24,4 The rotating vector representation of SHM(旋转矢量表示法),A vector with a length of A is rotating about point O at an angular velocity (see in Fig).,25,The projection(投影) P of this rotating vector in x-axis is given by,which is as same as the equation of SHM.,A rotating vector,SHM,26,D

12、iscussions:,Determined the initial phase by the rotating vector representation.,(1)过平衡位置沿x轴正向运动;,(2)过平衡位置沿x轴负向运动;,27,10-2 The Energy of SHM,The potential energy of the system is:,and its kinetic energy is equal to:,1 Total energy of SHM,28,The total energy of SHM is constant.,(1)弹簧振子作简振动过程中机械能守恒。 (2

13、)对于一定的弹簧振子,谐振动的机械能与振幅平方成正比。振幅越大,振子的总机械能越大。,Considering k=m2, the total energy,Conclusions:,29,Average value of energy during a period,Conclusion:谐振动在一个周期内的平均势能和平均动能相等,均为kA2/4。,2 Average energy of SHM,30,3 Transformation of the energy of SHM,振动过程中,动能和势能大小时刻改变并且相互转化,但总能量保持不变。,Conclusions:,31,Example(1

14、):,A mass of 100g vibrates horizontally in SHM with a frequency of 20Hz and an amplitude of 15cm. Calculate: (1)The total energy of the motion. (2)The velocity of the mass when it is 10cm from the center point.,32,Solution:,33,Example(2):,P15.例10.6、10.7,34,10-3 Damped Vibration& Forced Vibration Res

15、onance (阻尼振动 受迫振动 共振),1. Damped Vibration,The body is subject to the damping force, and its amplitude A and energy E will decrease to zero gradually.振动的物体在实际运动过程中,会受到阻碍其运动的力的作用,能量逐渐受到损失,振幅逐渐衰减的振动称为阻尼振动或减幅振动。,Fig. 10-7,35,2. Forced Vibration(受迫振动),The system is subject to a periodic external force(系统

16、受周期性强迫力作用下的振动),3. Resonance(共振),Phenomena that an maximum of amplitude of a damping vibration appears.,36,10-4 Electromagnetic Oscillation,1. LC circuit,charge,Fig. 10-8,37,2. Oscillation of LC circuit,Fig. 10-9,38,10-5 Superposition of two SHMs,1. The composition of two harmonic vibrations with the

17、 same direction and frequency.,The displacements resulting from two harmonic vibrations in the same straight line (x-axis) are:,39,It can be proved that the displacement of composition vibration is given by:,with the same angular frequency .,It is easy to prove by using the rotating vectors:,40,Spec

18、ial examples:,(1)when 2-1=2k, A=A1+A2, A reaches maximum(Enhancing) (2)when 2-1=2k, A=|A1-A2|, A reaches minimum(Weakening) (3)Generally, |A1- A2|AA1+A2,41,The two harmonic vibrations are,Find their composition vibration.,Example:,Plot the rotating vectors of x1,x2,x=x1+x2.,Solution:,1=/2, 2=0.,42,T

19、hen:,Hence:,43,2. The composition of two harmonic vibrations with the same direction and different frequencies.,10-5 Superposition of two SHMs,The total amplitude will varies periodically with the time,whose Chinese name is 拍.,3. The composition of two harmonic vibrations with the same frequency and

20、 vertical directions.,The orbit of the end of the composite vibration vector will be a line, a ellipse or a circle, which depends on the phase difference of the two vibrations.,Applications:polarized light 偏振光,44,10-5 Superposition of two SHMs,4. The composition of two harmonic vibrations with verti

21、cal directions and different frequencies.,The orbit of the end of the composite vibration vector will be complex shapes. An important and typical case of them is called lissajous figures.(李萨如图形),It is proved that any complex vibration can be decomposited into many SHMs. SHM is the simplest and basic vibration.,45,习题课,Example(1):,轻弹簧原长l0,平衡时伸长了x0,现手拉下L后无初速释放,请写出物体的运动方程。,46,Solution:,47,The simple pendulum(单摆),Prove: The motion of the simple pendulum is SHM.,Example(2):,

展开阅读全文
相关资源
猜你喜欢
相关搜索

当前位置:首页 > 企业管理 > 管理学资料

本站链接:文库   一言   我酷   合作


客服QQ:2549714901微博号:道客多多官方知乎号:道客多多

经营许可证编号: 粤ICP备2021046453号世界地图

道客多多©版权所有2020-2025营业执照举报