题目内容
某同学在做“利用单摆测重力加速度”的实验中,先测得单摆的摆长
,然后用秒表记录了单摆振动
次所用的时间为
,进而算出周期
。
(1)该同学测得的
值偏小,可能原因是:
(2)为了提高实验精度,在实验中可改变几次摆长
并测出相应的周期
,从而得出一组对应的
和
的数值,再以
为横坐标、
为纵坐标将所得数据连成直线,并求得该直线的斜率
。则重力加速度
。(用
和有关常量表示)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929087285.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929227304.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929243271.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929259309.png)
(1)该同学测得的
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929290278.png)
A.开始计时时,秒表过迟按下 |
B.摆线上端未牢固地系于悬点,振动中出现松动,使摆线长度增加了 |
C.实验中误将51次全振动计为50次 |
D.实验中误将49次全振动计为50次 |
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929087285.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929259309.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929087285.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929259309.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929087285.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929695368.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929711320.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929820294.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929711320.png)
BC
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929883573.png)
分析:(1)根据重力加速度的表达式,分析g值偏小可能的原因.
(2)由重力加速度的表达式,根据数学知识分析T2-l图线斜率的意义.
解答:解:(1)A、开始计时时,秒表过迟按下,测得的单摆周期变小,根据g=
可知,测得的g应偏大.故C错误.
B、摆线上端未牢固地系于悬点,振动中出现松动,使摆线长度增加了,测得的单摆周期变大,根据g=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929945618.png)
可知,测得的g应偏小.故B正确.
C、实验中误将51次全振动计为50次,根据T=
求出的周期变大,g偏小.故C正确.
D、实验中误将49次全振动计为50次,根据T=
求出的周期变小,g偏大.故D错误.
故选BC
(2)根据重力加速度的表达式g=
可知,T2-l图线斜率k=
,则g=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124930226556.png)
故答案为:(1)BC;(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124930226556.png)
点评:单摆的周期采用累积法测量可减小误差.对于测量误差可根据实验原理进行分析.图线可利用数学知识分析其物理意义.
(2)由重力加速度的表达式,根据数学知识分析T2-l图线斜率的意义.
解答:解:(1)A、开始计时时,秒表过迟按下,测得的单摆周期变小,根据g=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929945618.png)
B、摆线上端未牢固地系于悬点,振动中出现松动,使摆线长度增加了,测得的单摆周期变大,根据g=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929945618.png)
可知,测得的g应偏小.故B正确.
C、实验中误将51次全振动计为50次,根据T=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124930023393.png)
D、实验中误将49次全振动计为50次,根据T=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124930023393.png)
故选BC
(2)根据重力加速度的表达式g=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124929945618.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124930179542.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124930226556.png)
故答案为:(1)BC;(2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124930226556.png)
点评:单摆的周期采用累积法测量可减小误差.对于测量误差可根据实验原理进行分析.图线可利用数学知识分析其物理意义.
![](http://thumb2018.1010pic.com/images/loading.gif)
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