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1、1,第五讲 机 械 波,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,声波、水波、电磁波都是常见的波。,各种类型的波有其特殊性，但也有普遍的共性，可以用相同的数学形式来描述。,波动的分类：机械波、电磁波、物质波。,第五讲 机 械 波,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyrigh
2、t 2004-2011 Aspose Pty Ltd.,一、机械波产生的条件（源和路）,1.波源,2.连续介质, 1 机械波的形成和传播,二、机械波的类型,1.横波：介质中质点振动的方向与波的传播方向垂直,2.纵波：介质中质点振动的方向与波的传播方向平行,1)横波只能在固体中。,2)纵波在所有物质中都可以传播。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,波速,注意,1、有些波既不是横波也不是纵波。如：水表
3、面的波既非横波又非纵波。水波中的质元是做圆（或椭圆）运动的。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,flash,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,6,波动是振动状态（相位）的传播，是能量的传播 ，质元并未
4、“随波逐流” 。,后面质点的振动规律与前面质点的振动规律相同，只是位相上有一个落后。,重要结论:,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,三、波长、波的周期和频率 波速波动描述,1、波长l 同一时刻，两个相邻的相位差 为2的振动质点间的距离。,3、频率n 单位时间内波向前传播的完整波 的数目. (1s内向前传播了几个波长）,2、波的周期T 波传过一个波长的时间。,周期或频率只决定于波源的振动,Evalu
5、ation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,在波动过程中，某一振动状态在单位时间内传播的距离。,4、波速：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,平面波,球面波,各向同性均匀介质中，波线恒与波面垂直。,沿波线方向各质点的振动
6、相位依次落后。,四、波线和波面,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,五、简谐波,简谐波是最基本的波动，波源和介质 的各质点均作简谐振动。数学上的傅里叶 分解法告诉我们，任何非简谐振动都可以 分解为许多简谐振动的组合。因此，研究 简谐波特别重要。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.
7、0. Copyright 2004-2011 Aspose Pty Ltd.,一、平面简谐波的波函数,1、右行波的波动方程：,（1）已知O点振动表达式：,P点在t时刻的振动状态, 与O点在(t -x/u)时刻的振动状态相同。, 2 平面简谐波的波动方程,P点的相位比O点相位落后 。,参考点O,（滞后效应）,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,（2）如图，已知 P 点的振动方程：,思考,Evaluat
8、ion only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,（1）已知O点振动表达式：,思考,2、左行波的波动方程：,（超前效应）,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,（2）如图，已知 P 点的振动方程：,思考,Evaluation only
9、. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,注意：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,1、t 一定时的波形图 （快照）,讨论各质点在给定时刻的振动方向,二、波函数的物理意义,Evaluation only. Created with Asp
10、ose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例1：如图所示，为一简谐波在t=0时刻的波 形图。试写出O、1、2、3点的初位相,解：画出下一时刻的波形图,（1）,（2）,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,讨论质点在某一时刻的振动方向,质点的振动速度,2、x一定，t 变化。,表示
11、 点处质点的振动方程（ 的关系）,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,波速即波形平移的速度！,3、t、x都变化时：,对应跑动的波形（行波）,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,通常：,Evaluation
12、 only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例2：一平面简谐波某时刻的波形图如下，则OP之间的距离为多少厘米。,解题思路：,设波向右传播（P点落后于O点),思考：若设波朝左传，则,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例3：一平面波在
13、介质中以速度 u=20m/s 沿x 轴正方向传播，已知在传播路径上某点a点的振动方程为 ya=3cos4t (m) （1）以a为坐标原点写出波动方程。 （2）以距a点5m处的o点为坐标原点， 写出 波动方程。 (3) 求出oc 两点之间的位相差,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,解： ya=3cos4t,（1） a为坐标原点：,(2) 0点为坐标原点：,0 点振动方程,Evaluation onl
14、y. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(3) 求出oc 两点之间的位相差,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例4：沿X轴正方向传播的平面简谐波、在 t=2s 时的波形如图，已知A、u及 ，问：1）原点O的初相及P点的初相各为多大？2
15、）写出波动方程。,解题思路：,思考:求O、P两点之间的位相差。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,26,设有一行波：,质量为 的媒质元其动能为：,3 波的能量,质元的速度：,一、媒质中振动动能,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 As
16、pose Pty Ltd.,27,二、媒质中振动势能,可以证明：,势能：,总能：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd., 3 波的能量（补充）,胡克定律： 在弹性限度内，应力和应变成正比。,式中E为关于线变的比例系数，它随材料的不同而不同，叫杨氏模量。,补充：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile
17、 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,振动动能,以固体棒中传播的纵波为例分析波的能量.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,弹性势能,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose
18、 Pty Ltd.,能量极小,能量极大,体积元在平衡位置时，动能、势能和总机械能均最大.,体积元的位移最大时，动能、势能和总机械能均最小.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,波的能量密度：,平均能量密度：,随时间变化,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright
19、2004-2011 Aspose Pty Ltd.,也适用于球面波,平均能流,三、 能流和能流密度,1、能流 单位时间内通过某一截面的能量。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,2、能流密度（波的强度 I） 通过垂直于波速方向的单位面积的平均能流。,波强是矢量，其方向与波速方向相同。,波强是与振幅的平方成正比，其单位是W/m2。,Evaluation only. Created with Aspos
20、e.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例：如图，某一点波源发射功率为40瓦，求球面波上单位面积通过的平均能流。,解：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,（1）在均匀不吸收能量的媒质中传播的平面波在行进方向上振幅不变。,证明：因为 所以在单位时间内通过 和 面的能量应该相等,
21、所以,平面波振幅相等。,平面波和球面波的振幅,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,设距波源单位距离的振幅为A，则距波源 r 处的振幅为,（2）球面波振幅与它离波源的距离成反比。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty
22、Ltd.,一、惠更斯原理：波阵面(波前)上的每一点，都是发射子波的新波源，其后任意时刻，这些子波的包络面就是新的波阵面。,用惠更斯原理解释波的传播行为, 4 惠更斯原理 波的叠加和干涉,视频,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,波的衍射,水波通过狭缝后的衍射,波在传播过程中遇到障碍物时，能绕过障碍物的边缘，在障碍物的阴影区内继续传播.,惠更斯原理解释波的衍射：,Evaluation only. Cr
23、eated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,如你家在大山后,听广播和看电视哪个更容易? (若广播台、电视台都在山前侧),Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,1、若有几列波同时在介质中传播，则它们各自将以原有的振幅、频率和波长独立传播；并不因为其它波的
24、存在而改变。 （独立性）,2、在几列波相遇处，质元的位移等于各列波单独传播时在该处引起的位移的矢量和 （叠加性）,二、波的叠加原理：,细雨绵绵 独立传播,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,能分辨不同的声音正是这个原因； 叠加原理的重要性在于可以将任一复杂的波分解为简谐波的组合。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Cli
25、ent Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,三、波的干涉,相干条件:,相同的频率 恒定的初相位差 振动方向相同,干涉现象两列波相遇区域内，某些点的振动始终加强，另一些点的振动始终减弱的现象,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,传播到 P 点引起的振动：,波源S1:,波源S2:,Evaluation only. Created wi
26、th Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,1 ) 合振动的振幅（波的强度）在空间各点的分布随位置而变，但是稳定的.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,称 为波程差,干涉加强,干涉减弱,如图示，,当两相干波源为同相波源,即 时,Evaluation only. C
27、reated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,解：,例1：如图S1、S2为两平面简谐波相干波源，S2的位相比S1的位相超前 ， S1在P点引起的振幅为0.3m，S2在P点引起的振幅为0.2m，求P点的合振幅。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Ev
28、aluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,解：S1、S2为两个相干波 源，且初位相相同。,加强：,加强点的分布：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,减弱：,减弱点的分布：,Evaluation only. Cre
29、ated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,一、 产生驻波的演示实验, 5 驻 波,现象： AB弦线被分成几段长度相等的作稳定振动的部分，有些点始终静止，另一些点振动最强。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Crea
30、ted with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,驻波是入射波与反射波干涉的结果。,说明:,二、 驻 波方 程,设：,求出驻波的表达式：,合振动的振幅项,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,（1）各点都在作简谐振动，振幅：,（2）波节位置：,讨论：,相邻波节（
31、腹）间距：,波腹位置：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,3）相邻两波节之间质点振动同相位，任一波节两侧振动相位相反，在波节处产生 的相位跃变 .（与行波不同，无相位的传播）.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty
32、 Ltd.,例1：如图，若o、 处分别有两个相干波源， 其振动方程分别为：,求波腹和波节的位置。,解题思路：,驻波,右行波,左行波,对其中的任一点 x,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,当波由波密煤质入射到波疏煤质，在反射点，入射波和反射波的位相相同（即无半波损失）,波疏煤质,三、相位跃变（半波损失）,波密煤质,界面,入射波在界面上反射时，引起相位相反（相位有p跃变）,有半波损失动画,无半波损失动
33、画,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,解：（1）取O点为坐标原点，则入射波的波动方程为：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,反射波的波动 方程为：,Evaluation only. Created w
34、ith Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(2)能形成驻波的两列相干波，传播方向相反，已知入射波的波动方程为,则反射波的波动方程必可设为,在B点，xl1处是波节,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,解题思路：,Evaluation only. Created w
35、ith Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例4：在弹性媒质中有一沿X轴正向传播的平面波，其波动方程为： 若在X5.00m处有一媒质分解面，且在分解面处位相突变 ，设反射波的强度不变，试写出反射波的波动方程。,解题思路：,由题意知，在X5.00m处是一波节，,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011
36、Aspose Pty Ltd.,6 多普勒效应,当鸣笛的火车开向站台，站台上的观察者听到的笛声变尖，即频率升高；相反，当火车离开站台，听到的笛声频率降低。,现象,多普勒效应：观察者接受到的频率有赖于波源或观察者运动的现象。,视频,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,接收频率单位时间内观测者接收到的振动次数或完整波数.,人耳听到的声音的频率与声源的频率相同吗？,只有波源与观察者相对静止时才相等.,Ev
37、aluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,一、机械波的多普勒现象,1、声源不动,设观察者（observer)相对于媒质的运动速度为,（2）人远离声源运动,（1）人向声源运动,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,二、
38、 观察者不动,波源相对介质以 运动,动画,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,设声源（sourcer)相对于媒质的运动速度为,则：,（1）声源向观察者移动,则：,（2）声源背离观察者移动,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose
39、 Pty Ltd.,观察者向波源运动 + 远离 -,波源向观察者运动 - 远离 +,三 波源与观察者同时相对介质运动,动画,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,不论是波源运动，还是观察者运动，或是两者同时运动，定性地说：,结论：,两者靠近，接受到的频率高于原来波源的频率； 两者远离，接受到的频率低于原来波源的频率；,应作为运动速度沿波源和观察者连线方向的分量。,Evaluation only. Cr
40、eated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,当 时，所有波前将聚集在一个圆锥面上，波的能量高度集中形成冲击波或激波，如核爆炸、超音速飞行等.,1、报警和监测车速。医学上，测量血流速度、做超声心动等。,三、应用,2、跟踪人造地球卫星等。,3、天文学家利用电磁波红移说明大爆炸理论；,4、用于贵重物品、机密室的防盗系统；,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client P
41、rofile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,冲击波（激波),当波源向接收器运动时，接收器接收的频率比波源的频率大，但当波源的速度等于波的速度，波源总在波阵面上。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,如果波源的速度大于波的速度，波源总在波阵面前面。,冲击波,马赫角,马赫数=,飞机冲破声障时将发出巨大声响，造成噪声污染,视频,Evaluatio
42、n only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,警察用多普勒测速仪测速,超声多普勒效应测血流速,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例1：一静止波源向一飞行物发射n=30KHZ的超声波，飞行物离波源而去，在波源处测得发射波与反射波拍频
43、为Dn=100Hz。已知声速为u=340m/s,计算飞行物离去速度的大小。,解：,飞行物得到的频率为：,波源处测得飞行物的反射波的频率为：,拍频：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例2：如图，A、B为两个汽笛，其频率均为500Hz，A为静止的，B以60m/s的速率向右运动。在两个汽笛之间有一观察者R，以30m/s的速度也向右运动，已知空气中的声速为330m/s，求： （1）观察者听到来自A的频率
44、。 （2）观察者听到来自B的频率。 （3）观察者听到的拍频。,解：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,二、光的多普勒效应*,光的传播是“自己到达”，不需媒质传播，光速不变，C=nl,光源与观察者的运动是相对的，若相对速度为V，则可用相对论证明光波的频率变为：,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,红移：当光源远离接收器时，接收到的频率变小，因而波长变长。 如来自星球与地面同一元素的光谱比较，发现几乎都发生红移。这就是 “大爆炸”宇宙学理论的重要依据。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,素材和资料部分来自网络，如有帮助请下载!,