题目内容
B(选修模块 3 一4)( l )一个摆长为 L 的单摆做简谐运动,若从某时刻开始计时(取t=0),当 t=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_ST/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_ST/images2.png)
( 2 )如图2所示为一列简谐波在 tl=0 时刻的图象,此时波中质点 M 的运动方向沿y轴正方向,到 t2=0.65s 时质点 M 恰好第 4 次到达y轴正方向最大位移处,则该波的传播方向为______,波速为______m/s.
( 3 )一组平行的细激光束,垂直于半圆柱玻璃的平面射到半圆柱玻璃上,如图3所示.已知光线 I 沿直线穿过玻璃,它的入射点是O,光线Ⅱ的入射点为 A,穿过玻璃后两条光线交于一点.已知玻璃截面的圆半径为 R,0A=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_ST/2.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_ST/3.png)
【答案】分析:(1)根据单摆的周期公式知,t=![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/0.png)
时,经历了
,摆球具有负向的最大加速度,知处于正向最大位移处,根据该规律确定单摆的位移时间图线.
(2)根据质点的振动方向,通过上下坡法确定波的传播方向.根据质点 M 恰好第 4 次到达y轴正方向最大位移处,求出质点的周期,通过波长求出波速的大小.
(3)作出光路图,结合折射定律和几何关系求出两条光线射出玻璃后的交点与O点的距离.
解答:解:(1)单摆的周期T=
,t=![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/4.png)
时,经历了
,摆球具有负向的最大加速度,知处于正向最大位移处,可知零时刻处于平衡位置,且向负方向运动,故B正确,A、C、D错误.
故选B.
(2)波中质点 M 的运动方向沿y轴正方向,根据上下坡法,知波的传播方向为+x方向.到 t2=0.65s 时质点 M 恰好第 4 次到达y轴正方向最大位移处,因为经过
第一次到达y轴正向最大位移处,则有3
,解得T=0.2s.则波速v=
.
(3)两条光线的光路图如图所示,设射出玻璃后的交点是P,光线Ⅱ从玻璃射出时的入射角为i,折射角为r,根据折射定律得,![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/images10.png)
n=![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/10.png)
由几何关系可得i=30°
代入r=60°
由几何关系可得:OP=2Rcos30°=
.
故答案为:(1)B (2)+x方向,2m/s (3)![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/12.png)
点评:本题考查了振动和波动的关系以及几何光学问题,难度不大,关键要熟悉教材,理解基本规律.
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/2.png)
(2)根据质点的振动方向,通过上下坡法确定波的传播方向.根据质点 M 恰好第 4 次到达y轴正方向最大位移处,求出质点的周期,通过波长求出波速的大小.
(3)作出光路图,结合折射定律和几何关系求出两条光线射出玻璃后的交点与O点的距离.
解答:解:(1)单摆的周期T=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/6.png)
故选B.
(2)波中质点 M 的运动方向沿y轴正方向,根据上下坡法,知波的传播方向为+x方向.到 t2=0.65s 时质点 M 恰好第 4 次到达y轴正方向最大位移处,因为经过
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/9.png)
(3)两条光线的光路图如图所示,设射出玻璃后的交点是P,光线Ⅱ从玻璃射出时的入射角为i,折射角为r,根据折射定律得,
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/images10.png)
n=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/10.png)
由几何关系可得i=30°
代入r=60°
由几何关系可得:OP=2Rcos30°=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/11.png)
故答案为:(1)B (2)+x方向,2m/s (3)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203103863936059/SYS201310282031038639360012_DA/12.png)
点评:本题考查了振动和波动的关系以及几何光学问题,难度不大,关键要熟悉教材,理解基本规律.
![](http://thumb2018.1010pic.com/images/loading.gif)
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