题目内容
(8分)一水平弹簧振子做简谐运动,其位移和时间关系如图所示.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241322436132855.png)
(1)求t=0.25×10-2 s时的位移.(2)从t=0到t=8.5×10-2 s的时间内,质点的路程、位移各为多大?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241322436132855.png)
(1)求t=0.25×10-2 s时的位移.(2)从t=0到t=8.5×10-2 s的时间内,质点的路程、位移各为多大?
(1).x=-1.41cm,(2).路程=0.34m 位移=0
简谐运动的表达式为x=Asin(ωt+φ),由图可知表达式为x=2sin(100πt-
)cm
把t=0.25×10-2 s代入表达式可得位移为![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132243675753.png)
质点每经过一个周期运动路程为4个振幅8cm,质点振动周期为2×10-2 s,
从t=0到t=8.5×10-2 s的时间内,共有
个周期,所以质点的路程为4×8+2=34cm,将t=8.5×10-2 s代入运动表达式可知位移为0
故答案为:(1).x=-1.41cm,(2).路程=0.34m 位移=0
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132243644421.png)
把t=0.25×10-2 s代入表达式可得位移为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132243675753.png)
质点每经过一个周期运动路程为4个振幅8cm,质点振动周期为2×10-2 s,
从t=0到t=8.5×10-2 s的时间内,共有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824132243706366.png)
故答案为:(1).x=-1.41cm,(2).路程=0.34m 位移=0
![](http://thumb2018.1010pic.com/images/loading.gif)
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