题目内容
有一单摆,其摆长l="1.02" m,摆球的质量m="0.10" kg,已知单摆做简谐运动,单摆振动30次用的时间t="60.8" s,试求:
(1)当地的重力加速度是多大?
(2)如果将这个摆改为秒摆,摆长应怎样改变?改变多少?
(1)当地的重力加速度是多大?
(2)如果将这个摆改为秒摆,摆长应怎样改变?改变多少?
(1)9.79 m/s2.(2)缩短0.027 m.
(1)当单摆做简谐运动时,其周期公式
T=2π
,由此可得g="4π" 2l/T2,只要求出T值代入即可.
因为T=
=
s="2.027" s,
所以g=4π2l/T2=(4×3.142×1.02)/2.0272 m/s2="9.79" m/s2.
(2)秒摆的周期是2 s,设其摆长为l0,由于在同一地点重力加速度是不变的,根据单摆的振动规律有:
=
,
故有:l0=
=
m="0.993" m.
其摆长要缩短Δl=l-l0="1.02" m-0.993 m="0.027" m.
T=2π
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120010308300.gif)
因为T=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120010323231.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120010339351.gif)
所以g=4π2l/T2=(4×3.142×1.02)/2.0272 m/s2="9.79" m/s2.
(2)秒摆的周期是2 s,设其摆长为l0,由于在同一地点重力加速度是不变的,根据单摆的振动规律有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120010339252.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120010355301.gif)
故有:l0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120010370410.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120010386451.gif)
其摆长要缩短Δl=l-l0="1.02" m-0.993 m="0.027" m.
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目