当前位置:文档之家› 2017-2018学年高中物理教科版选修3-4课时跟踪检测:(四) 阻尼振动 受迫振动 Word版含解析

2017-2018学年高中物理教科版选修3-4课时跟踪检测:(四) 阻尼振动 受迫振动 Word版含解析

2017-2018学年高中物理教科版选修3-4课时跟踪检测:(四) 阻尼振动 受迫振动 Word版含解析
2017-2018学年高中物理教科版选修3-4课时跟踪检测:(四) 阻尼振动 受迫振动 Word版含解析

课时跟踪检测(四)阻尼振动受迫振动

1.下列说法正确的是()

A.实际的自由振动必然是阻尼振动

B.在外力作用下的振动是受迫振动

C.阻尼振动的振幅越来越小

D.受迫振动稳定后的频率与自身物理条件有关

2.下列说法正确的是()

A.只有受迫振动才能发生共振现象

B.一切振动都能发生共振现象

C.只要振动受到阻力,它一定做阻尼振动

D.若振动受到阻力,它也可能做无阻尼振动

3.在飞机的发展史中有一个阶段,飞机上天后不久,飞机的机翼(翅膀)很快就抖动起来,而且越抖越厉害。后来人们经过了艰苦的探索,利用在飞机机翼前缘处装置一个配重杆的方法,解决了这一问题。在飞机机翼前装置配重杆的目的主要是()

A.加大飞机的惯性

B.使机体更加平衡

C.使机翼更加牢固

D.改变机翼的固有频率

4.一砝码和一轻弹簧构成弹簧振子,如图1所示的装置可用于研究该弹簧振子的受迫振动。匀速转动把手时,曲杆给弹簧振子以驱动力,使振子做受迫振动。把手匀速转动的周期就是驱动力的周期,改变把手匀速转动的速度就可以改变驱动力的周期。若保持把手不动,给砝码一向下的初速度,砝码便做简谐运动,振动图线如图2甲所示。当把手以某一速度匀速转动,受迫振动达到稳定时,砝码的振动图线如图乙所示。

图1

图2

若用T0表示弹簧振子的固有周期,T表示驱动力的周期,Y表示受迫振动达到稳定后砝码振动的振幅,则()

A.由图线可知T0=4 s

B.由图线可知T0=8 s

C.当T在4 s附近时,Y显著增大;当T比4 s小得多或大得多时,Y很小

D.当T在8 s附近时,Y显著增大;当T比8 s小得多或大得多时,Y很小

5.把一个筛子用四根弹簧支起来,筛子上安一个电动偏心轮,它每转一周,给筛子一个驱动力,这样就做成了一个共振筛,如图3所示。筛子做自由振动时,完成10次全振动用时15 s,在某电压下,电动偏心轮转速是36 r/min。已知增大电压可使偏心轮转速提高;增加筛子质量,可以增大筛子的固有周期,那么要使筛子的振幅增大,下列做法正确的是()

图3

A.提高输入电压B.降低输入电压

C.增加筛子质量 D. 减小筛子质量

6.一洗衣机的脱水筒在正常工作时非常平稳,当切断电源后,发现洗衣机先是振动越来越剧烈,然后振动在逐渐减弱。对这一现象,下列说法中正确的是()

①正常工作时,洗衣机脱水筒的运转频率比洗衣机的固有频率大;②正常工作时,洗衣机脱水筒的运转频率比洗衣机的固有频率小;③正常工作时,洗衣机脱水筒的运转频率等于洗衣机的固有频率;④当洗衣机振动最剧烈时,洗衣机脱水筒的运转频率恰好等于洗衣机的固有频率。

A.①②B.②③

C.①④ D. ②④

7.图4是探究单摆共振条件时得到的图像,它表示振幅跟驱动力频率之间的关系,(g=9.8 m/s2)请回答:

图4

(1)这个单摆的摆长是多少?

(2)如果摆长变长一些,画出来的图像的高峰将向哪个方向移动?

8.如图5所示,在曲轴A上悬挂一个弹簧振子,如果转动把手,曲轴可以带动弹簧振子上下振动。问:

图5

(1)开始时不转动把手,而用手往下拉振子,然后放手让振子上下振动,测得振子在10 s 内完成20次全振动,振子做什么振动?其固有周期和固有频率各是多少?若考虑摩擦和空气阻力,振子做什么振动?

(2)在振子正常振动过程中,以转速4 r/s匀速转动把手,振子的振动稳定后,振子做什么运动?其周期是多少?

(3)若要振子振动的振幅最大,把手的转速应多大?

答 案

1.选AC 实际的自由振动,必须不断克服外界阻力做功而消耗能量,振幅会逐渐减小,必然是阻尼振动,故A 、C 正确;只有在周期性外力(驱动力)的作用下物体所做的振动才是受迫振动,B 错;受迫振动稳定后的频率由驱动力的频率决定,与自身物理条件无关,D 错。

2.选AD 发生共振的条件是f 固=f 驱,所以A 对、B 错。无阻尼振动是振幅不变的振动,当受阻力时,若外界补充能量,则也可能做无阻尼振动,C 错,D 对。

3.选D 飞机抖动得厉害是因为发生了共振现象,想要解决这一问题需要使系统的固有频率与驱动力的频率差别增大,在飞机机翼前缘处装置一个配重杆,改变的是机翼的固有频率,故选项D 正确。

4.选AC 该题是考查机械振动图像和共振现象的题目。细读图中所给情景,可知:第1幅图是描述弹簧振子在不受驱动力的情况下振动情况的图像,此时的振动周期是该弹簧振子的固有周期,从图中可直接读出固有周期为4 s ,A 正确,B 错误;当驱动力的频率与固有频率相近时,发生共振,振幅显著增大,所以当曲杆的转动周期与弹簧振子的固有周期4 s 接近的时候,振幅Y 显著增大,其他情况下Y 很小,C 正确,D 错误。

5.选AC 在题给条件下,筛子振动的固有周期T 固=1510

s =1.5 s ,电动偏心轮的转动周期(对筛子来说是驱动力的周期)T 驱=6036

s =1.67 s 。要使筛子振幅增大,就得使这两个周期值靠近,可采用两种做法:第一,提高输入电压使偏心轮转得快一些,减小驱动力的周期;第二,增加筛子的质量使筛子的固有周期增大。正确选项为A 、C 。

6.选C 正常工作时,洗衣机脱水筒运转频率比断电后的频率高,共振发生在断电之后,所以正常工作时的频率比洗衣机的固有频率大。洗衣机切断电源,脱水筒的转动逐渐慢下来,在某一小段时间内洗衣机发生了剧烈的振动,说明了此时脱水筒的运转频率与洗衣机的固有频率相同,发生了共振。此后脱水筒转速减慢,则f 驱<f 固,所以共振现象消失,洗衣机的振动随之减弱,故C 正确。

7.解析:(1)从共振曲线知,单摆的固有周期

T =1f 固=10.3

s =103 s , 因为T =2π l g , 所以l =gT 24π2=9.8×????10324×3.142

m ≈2.76 m 。 (2)若摆长变长一些,则由周期公式T =2π l g

可知,单摆固有周期增大,则固有频率

变小,因此图像的高峰将向原点方向移动。

答案:(1)2.76 m (2)向原点方向移动

8.解析:(1)根据题意振子做自由振动,

T 固=t n =1020 s =0.5 s ,f 固=1T 固=10.5

Hz =2 Hz 。 由于摩擦力和空气阻力的存在,振子克服摩擦力和阻力做功消耗能量,使其振幅越来越小,故振动为阻尼振动。

(2)振子做受迫振动。由于把手转动的转速为4 r/s ,即驱动力频率为f 驱=4 Hz ,周期T 驱=0.25 s 。弹簧振子振动达稳定状态后,其周期等于驱动力的周期,T =T 驱=0.25 s 。

(3)要使弹簧振子的振幅最大,处于共振状态,必须使驱动力的频率f 驱等于它的固有频率f 固,即f 驱=f 固=2 Hz ,故把手的转速应为n =2 r/s 。

答案:见解析

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