机械振动第1章习题
机械振动复习题
机械振动复习题第⼀章⼀填空题1.机械运动是⼀种特殊形式的运动，在这种运动过程中，机械系统将围绕作运动。
2.从能量的⾓度看，惯性是保持的元素，恢复性是贮存的元素，阻尼是使能量散逸的元素。
3.⼀个质点在空间作⾃由运动，决定其位置需要个独⽴的坐标，⾃由度数为,⽽由n个相对位置可变的质点组成的质点系，其⾃由度数为，当系统收到r个约束条件时，系统的⾃由度数是。
4、系统的⾃由度是表明能够描述系统各部分在任⼀瞬时位置的独⽴的最⼩数⽬。
⼆、判断题1、简谐振动的加速度，其⼤⼩与位移呈正⽐，⽽⽅向与位移相反，始终指向平衡位置。
（）2、简谐运动是周期运动。
（）3、所有表⽰周期振动的周期函数都可以展开成Fourier级数的形式。
（）4、⼴义坐标必须能完整地描述系统的运动。
（）5、两个同频率的简谐振动在同⽅向的合成运动是该频率的简谐振动。
（）三、简答题1、构成机械振动系统的基本元素有哪些？其作⽤？四、选择题1、单摆的⾃由度是（）。
A、1B、2C、3D、02、⼀个单⾃由度系统都可以⽤这样⼀个理论模型来描述：它是由以下哪三个基本元件组成（）ωA.理想的弹簧kB.理想的阻尼cC.理想的质量mD.理想的固有频率nE.理想的阻尼⽐ξ五、名词解释1、⾃由度六、计算题P16 1-1、1-2⼀、⼀填空题填空题1、阻尼对抑制系统近旁的运动有决定作⽤，⽽对系统在⾮共振频率的运动影响不⼤。
2、冲击⼒的特点是数值很⼤，但作⽤时间很。
3、单位作⽤下单⾃由度系统的响应称为脉冲响应函数。
4、振动问题的全解由态解和瞬态解组成。
5、振动测试仪器有三种基本形式：测试、和的仪器。
它们都是根据引起系统振动的原理⼯作的。
5、对于线性系统，叠加原理成⽴，即各激励⼒共同作⽤所引起的系统稳态响应为各激励⼒（）时引起的系统各稳态响应的总和。
⼆、选择题1、对于机械系统有三种典型的强迫振动的情况（）A.系统本⾝的不平衡引起的强迫振动B.简谐激励⼒作⽤下强迫振动C.基础运动引起的强迫振动D⽀承运动引起的强迫振动三、判断题1、线性系统内各个激励产⽣的响应是互不影响的。
第一章机械运动习题答案
习题（一）一、选择题1. 一质点在xy 平面内运动，其运动方程为2,ct b y at x +==，式中a 、b 、c 均为常数。
当运动质点的运动方向与x 轴成45º角时，它的速率为[ B ]。
A ．a ；B ．a 2；C ．2c ；D ．224c a +。
2. 一质点以匀速率在xy 平面内运动，如图1-11所示。
则经轨道上的a 、b 、c 、d 四点时，质点的加速度最大的点是[ B ]。
A ．aB ．bC ．cD ．d3. 下列说法中正确的是（ D ）A ． 加速度恒定不变时，物体的运动方向也不变；B ．平均速率等于平均速度的大小；C. 当物体的速度为零时，加速度必定为零；D ．质点作曲线运动时，质点速度大小的变化产生切向加速度，速度方向的变化产生法向加速度。
4. 设木块沿光滑斜面从下端开始往上滑动，然后下滑，则表示木块速度与时间关系的 曲线（如图1-99所示）是[ D ]。
二、填空题1. 一质点沿x 轴运动，其运动方程为225t t x -+=（SI ）。
质点的初速度为 2m/s ，第4秒末的速度为 -6m/s ，第4秒末的加速度为-2m/s 2。
2. 质点作直线运动，其速度与时间的关系曲线如图1-100所示。
图中过A 点的一切线AC 的斜率表示 t 1 时刻加速度 ，割线AB 的斜率表示 t 1 时刻到t 2时刻的平均加速度 ，曲线下的面积()⎰21t t dt t v 表示 从t 1时刻到t 2时刻质点的位移 。
三、计算题1. 已知质点的运动方程为t x 2=，24t y -=（SI ）。
试求：（1）试导出质点的轨道方程，并图示质点的运动轨迹；（2）计算t=1s 和t=2s 时质点的位置矢量，并计算1s 到2s 之间质点的平均速度和位移；（3）计算质点在第2秒末时的速度和加速度，并说明质点作何种运动？答：(1)由2x t =得2xt = ，代入22y t =-224x =-即为轨道方程。
机械基础第一章练习题
第1章1。
单选题(本大题共94小题)1。
（1分)以零部件装备配或更换时不需要挑选、辅助加工与修配为条件的互换性，等于_______互换性。
A、绝对B、完全C、有限D、不完全2.（1分）分组装配法等于典型的________互换性，其方法是零件加工完后，根据零件实测尺寸的大小，将制成的零件分成若干组，使每组内的尺寸差别较小，然后对相应组的零件进行装配。
A、绝对B、不完全C、完全D、有限3.（1分）国际标准化组织的号是________。
A、ISOB、ANSIC、JISD、KS4.(1分）不属于国家标准的代号是_________。
A、GBB、GB/TC、JBD、QB5.（1分）__________℃为标准计量温度.A、20B、25C、0D、1006.（1分）关于孔和轴的概念，下列说法中错误的是_______.A、圆柱形的内表面为孔，圆柱形的外表面为轴B、孔和轴的形状一定都是圆的C、从装配关系上看，包容面为孔，被包容面为轴D、从加工过程上看,切削过程中尺寸由小变大的为孔,尺寸由大变小的为轴7。
(1分）基本尺寸是______.A、测量时得到的B、加工时得到的C、装备后得到的D、设计时给定的8。
(1分)最大极限尺寸与基本尺寸的关系是_______。
A、前者大于后者B、前者小于后者C、前者等于后者D、两者之间的大小无法确定9。
（1分)某尺寸的实际偏差为零，则实际尺寸_________.A、必定合格B、为零件的真实尺寸C、等于基本尺寸D、等于最小极限尺寸10。
（1分)当上偏差或下偏差为零时，在图样上进行标注时________。
A、必须标出B、不用标出C、标与不标皆可D、视具体情况而定11.（1分)极限偏差是________.A、设计时确定的B、加工后测量得到的C、实际尺寸减基本尺寸的代数差D、最大极限尺寸与最小极限尺寸之差12.(1分）关于尺寸公差，下列说法中正确的是_________。
A、尺寸公差只能大于零，故公差值前应标“+”号B、尺寸公差是用绝对值定义的,没有正、负的含义，故公差值不应标“+”号C、尺寸公差不能为负值,但可为零值D、尺寸公差为允许尺寸变动范围的界限值13.(1分)Φ200030 .0+mm与Φ2000072 .0+mm相比，其尺寸精确程度________.A、相同B、前者高，后者低C、前者低，后者高D、无法比较14.(1分）当孔的最大极限尺寸与轴的最小极限尺寸的代数差为正值时,此代数差称为_____。
机械振动考题
机械振动考题第一章1.21.If energy is lost in any way during vibration, the system can be considered to be damped. (T)2.Superposition principle is valid for both linear and nonlinear systems(F)3.The frequency with which an initially disturbed system vibrates on its own is known as natural frequency(T)4.Any periodic function can be expanded into Fourier series(T)5.Harmonic motion is a periodic motion(T)6.The equivalent mass of several masses at different locations can be found using the equivalence of kinetic energy(T)7.The generalized coordinates are not necessarily Cartesian coordinates. (T)8.Discrete systems are same as lumped parameter systems(T)9.Consider the sum of harmonic motions,, withand The amplitude A is given by 30.8088(T)10.Consider the sum of harmonic motions, , withand The phase angle α is given by 1.57 rad. (F)第二章2.21.The amplitude of an undamped system will not change with time.(T)2.A system vibrating in air can be considered as a damped system(T)3.The equation of motion of a single degree of freedom system will be the same whether the mass moves in a horizontalplane or an inclined plane.(T)4.When a mass vibrates in a vertical direction, its weight can always be ignored in deriving the equation of motion(F)5.The principle of conservation of energy can be used to derive the equation of motion of both damped and undamped systems(F)6.The damped frequency can be larger that the undamped natural frequency of the system in some cases(F)7.The damped frequency can be zero in some cases. (T)8.The natural frequency of vibration of torsional system is given by where k and m denote the torsional spring constant and the polar mass moment of inertia, respectively(T)9.Rayleigh’s method is based on the principle of conservation of energy(T)10.The final position of the mass is always the equilibrium position in the case of Coulomb damping. (F)11.The undamped natural frequency of a system is given by , where is the static deflection of the mass(T)12.For an undamped system, the velocity leads the displacement by . (T)13.For an undamped system, the velocity leads the acceleration by (F)14.Coulomb damping can be called constant damping(T)15.The loss coefficient denotes the energy dissipated per radian per unit strain energy.(T)16.The motion diminishes to zero in both underdamped and overdamped cases. (T)17.The logarithmic decrement can be used to find the damping ratio(T)18.The hysteresis loop of the stress –strain curve of amaterial causes damping(T)19.The complex stiffness can be used to find the damping force in a system with hysteresis damping(T)20.The motion can be considered to be harmonic in the cases of hysteresis damping(T)第三章3.21.The magnification factor is the ratio of maximum amplitude and static deflection(T)2.The response will be harmonic if excitation is harmonic(T)3.The phase angle of the response depends on the system parameter m, c, k, and ω(T)4.The phase angle of the response depends on the amplitude of the forcing function.(F)5.During beating, the amplitude of the response builds up and then diminishes in a regular pattern (T)6.The Q-factor can be used to estimate the damping in a system (T)7.The half power points denote the values of frequency ratio where the amplification factor falls towhere Q is the Q-factor. (T)8.The amplitude ratio attains its maximum value at resonance in the case of hysteresis damping(F)9.The response is always in phase with the harmonic forcing function in the case of hysteresis damping(T)10.Damping reduces the amplitude ratio for all values of the forcing frequency. (T)11.The unbalance in a rotating machine causes vibration(T)12.The steady state solution can be assumed to be harmonicfor small values of dry friction force(T)13.In a system with rotational unbalance, the effect of damping becomes negligibly small at higher speeds. A set is a collection of objects(T)第四章4.21.The change in momentum is called impulse (T)2.The response of a system under arbitrary force can be found by summing the responses due toseveral elementary impulses (T)3.The response spectrum corresponding to base excitation is useful in the design of machinery subject to earthquakes (T)4.Some periodic functions can not be replaced by a sum of harmonic functions (F)5.The amplitudes of higher harmonics will be smaller in the response of a system. (T)6.The Laplace transform method takes the initial conditions into account automatically (T)7.The equation of motion can be integrated numerically even when the exciting force is nonperiodic (T)8.The response spectrum gives the maximum response of all possible single degree of freedom systems (T_9.For a harmonic oscillator, the acceleration and displacement spectra can be obtained from the velocity spectrum. (T)第五章5.21.The normal modes can also be called principal modes (T)2.The generalized coordinates are linearly dependent (F)3.Principal coordinates can be considered as generalizedcoordinates (T)4.The vibration of a system depends on the coordinate system (F)5.The nature of coupling depends in the coordinate system (T)6.The principal coordinates avoid both static and dynamic coupling.(T)7.The use of principal coordinates helps in finding the response of the system (T)8.The mass, stiffness, and damping matrices of a two degree of freedom system are symmetric (T)9.The characteristics of a two degree of freedom system are used in the design of dynamic vibration absorber (T)10.Semi-definite systems are also known as degenerate systems (T)11.A semi-definite system can not have non-zero natural frequencies (F)12.The generalized coordinates are always measured form the equilibrium position of the body (F)13.During free vibration, different degrees of freedom oscillate with different phase angles (F)14.During free vibration, different degrees of freedom oscillate at different frequencies (F)15.During free vibration, different degrees of freedom oscillate with different amplitudes (T)16.The relative amplitude of different degrees of freedom ina two degree of freedom system depend on the natural frequency (T)17.The modal vectors of a system denote the normal modes of vibration (T)第六章6.21.For a multidegree of freedom system, one equation of motion can be written for each degree of freedom (T) /doc/db13674645.html,grange’s equation cannot be used to derive the equations of motion of a multidegree of freedom system (F)3.The mass, stiffness, and damping matrices of a multidegree of freedom are always symmetric (T)4.The product of stiffness and flexibility matrices of a system is always an identity matrix (T)5.The modal analysis of a n-degree of freedom system can be conducted using r modes with r < n (T)6.For a damped multidegree of freedom system, all the eigenvalues can be complex (T)7.The modal damping ratio denotes damping in a particular normal mode (T)8.A multidegree of freedom system can have six of the natural frequencies equal to zero (T)9.The generalized coordinates will always have the unit of length (F)10.The generalized coordinates are independent of the conditions of constraint of the system (T)11.The generalized mass matrix of a multidegree of freedom system is always diagonal (F)12.The potential and kinetic energies of a multidegree of freedom system are always quadratic functions (T)13.The mass matrix of a system is always symmetric and positive definite (T)14.The stiffness matrix of a system is always symmetric andpositive definite (F)15.The rigid body mode is also called the zero mode. (T)16.An unrestrained system is also known as a semi-definite system. (T)17.Newton’s second law of motion can always be used to derive the equations of motion of a vibrating system (T) 第七章7.21.T he fundamental fr equency given by Durkerley’s formula will always be larger than the exact value (F)2.The fundamental frequency given by Rayleigh’s method will always be larger than the exact value (T)3.is a standard eigenvalue problem (F)4.is a standard eigenvalue problem (T)5.Jacobi method can find the eigenvalues of only symmetric matrices. (T)6.Jacobi method uses rotation matrices. (T)7.The matrix iteration method requires the natural frequencies to be distinct and well separated (T)8.In matrix iteration method, any computational error will not yield incorrect results (T)9.The matrix iteration method will never fail to converge to higher frequencies. (F)10.When Rayleigh’s method is used for a shaft carrying several rotors, the static deflection curve can be used as the appropriate mode shape. (T)11.Rayleigh’s method can be considered to be same as the conservation of energy for a vibrating system (T)第八章8.21.Continuous systems are same as distributed systems. (T)2.Continuous systems can be considered to have infinite number of degrees of freedom. (T)3.The governing equation of a continuous system is an ordinary differential equation. (F)4.The free vibration equations corresponding to the transverse motion of a string, the longitudinal motion of a bar and the torsional motion of a shaft have the same form. (T)5.The normal modes of a continuous system are orthogonal. (T)6.A membrane has zero bending resistance. (T)7.Rayleigh’s method can be considered as a method of conservation of energy.(T)8.Rayleigh-Ritz method assumes the solution as a series of functions that satisfy the boundary conditions of the problem. (T)9.For a discrete system, the boundary conditions are to be applied explicitly. (T)10.The Euler-Bernoulli beam theory is more accurate than the Timoshenko theory. (F)第九章9.21.Vibration can cause structural and mechanical failures. (T)2.The response of a system can be reduced by the use of isolators and absorbers (T)3.Vibration control means the elimination or reduction of vibration (T)4.The vibration caused by a rotating unbalanced disc can be eliminated by adding a suitable mass to the disc (T)5.Any unbalanced mass can be replaced by two equivalent unbalanced masses in the end planes of the rotor (T)6.The oil whip in the bearings can cause instability in a rotor system (T)7.The natural frequency of a system can be changed by varying its damping (F)8.The stiffness of a rotating shaft can be altered by changing the location of its bearings (T)9.All practical systems have damping. (T)10.High loss factor of a material implies less damping (F)11.Passive isolation systems require external power to function (F)12.The transmissibility is also called the transmission ratio. (T)13.The force transmitted to the foundation of an isolator with rigid foundation can never be infinity (F)14.Internal and external friction can cause instability in a rotating shaft at speeds above the first critical speed (T)。
机械振动一章习题解答
T = 2π
所以应当选择答案(C)。
m ( k1 + k 2 ) m = 2π k k1 k 2
习题 12—4
一质点作简谐振动，周期为 T，当它由平衡位置向 X 轴正方向运动 ］
时，从二分之一最大位移处到最大位移处这段路程所需要的时间为： ［ (A) T/4。 (B) T/12。 (C) T/6。 (D) T/8。
解： 单摆的振动满足角谐振动方程， 这里所给的 θ 是初始角位移，显然是从最大角位移处计时。由 旋转矢量法容易判断该单摆振动的初位相为 “0” ， 因此，应当选择答案(C) 。 −θm
题解 12―1 图
习题 12—2
轻弹簧上端固定，下端系一质量为 m1 的物体，稳定后在 m1 下边又
系一质量为 m2 的物体，于是弹簧又伸长了 ∆x ，若将 m2 移去，并令其振动，则 振动周期为： ［ (A) T = 2π ］
位相 ϕ = π 2 ，故振动方程为
x = 0.02 cos(1.5t +
π ) 2
(SI)
习题 12─17
两个同方向的简谐振动的振动方程分别为
1 ， x 2 = 3 × 10 − 2 cos 2π (t + ) 4
1 x1 = 4 × 10 − 2 cos 2π (t + ) 8
(SI)
求：合振动方程。 解：设合振动方程为
X
习题 12─12
一质点作简谐振动，振动图
线如图所示，根据此图，它的周期
4 O –2
2
t (s)
T=
ϕ=
，用余玄函数描述时的初位相
习题 12―12 图
。 解：根据振动图线可画出旋转矢量图，可得
t=2
∴ ∴
机械波一章习题解答
5m
习题 13―7 图
X(m)
y P = A cos(ω t + ϕ )
由所给的波形图容易得到： λ = 10 m ，A=0.10m，u=20m/s，而振动的圆频率
ω=
2πu 2π × 20 = = 4π rad/s λ 10
因为波是自左向右传播的， 由此可以判断出 P 点在 t=0 时刻正在最大位移一半处 且向 Y 轴负向运动，所以， P 点振动的初位相为 ϕ = π 3 。这样，P 处介质质点 的振动方程为
Y
B P C A PX
0 (B)
习题 13─11
如图所示，为一向右传
Y
0
X
播的简谐波在 t 时刻的波形图，BC 为 波密介质的反射面，波由 P 点反射。 则反射波在 t 时刻的波形图为： ［ ］ 解：因为 BC 为波密介质的反 射面，所以在反射时有“半波损失” ， 故反射波在 P 点引起的振动与入射波 在 P 点引的振动在位相上刚好相反，
(A) y P = 0.10 cos(4πt + π 3) 。 (B) y P = 0.10 cos(4πt − π 3) 。 (C) y P = 0.10 cos(2πt + π 3) 。 (D) y P = 0.10 cos(2πt + π 6) 。 解：设 P 点处的振动方程为
u=20m/s P
y0 = A cos(ω t ′ + φ )
由 t=3s 时的波形曲线可知 A = 2 × 10 −2 m , λ = 20 m，所以
ω = 2πν =
2πu π = rad/s λ 2
t ′ = 0 时，原点处质元处于负的最大位移处，则其位相为 φ = π ，所以，
故 x=0 处的振动方程为
2020_2021学年高中物理第一章机械振动4阻尼振动受迫振动
【核心归纳】 简谐运动、阻尼振动和受迫振动的比较:
振动类型 比较项目
产生条件
频率 振幅
简谐运动
不受阻力作用 固有频率 不变
阻尼振动
受阻力作用 频率不变 减小
受迫振动
受阻力和驱动力作用 驱动力频率
大小变化不确定
振动图像
形状不确定
实例
弹簧振子振动,单 敲锣打鼓发出的 扬声器纸盆振动发声、
摆做小角度摆动 声音越来越弱
4.阻尼振动 受迫振动
必备知识·自主学习
一、自由振动、阻尼振动 1.自由振动:系统___不__受__外__力____,也不受任何阻力,只在自身___回__复__力____作用下 的振动,又叫作___无__阻__尼__振__动____。 2.固有频率:___自__由____振动的频率。 3.阻尼:即阻力作用,通常包括___摩__擦__力____或其他阻力。 4.阻尼振动:振幅___逐__渐__减__小____的振动。
【情境思考】 如图为阻尼振动的振动图线,阻尼振动中,在振幅逐渐减小的过程中,振子的周 期如何变化?为什么?
提示:不变 周期与振幅无关
二、受迫振动与共振 1.驱动力:作用于振动系统的___周__期__性____的外力。 2.受迫振动:振动系统在___驱__动__力____作用下的振动。 3.受迫振动的频率:做受迫振动的频率等于___驱__动__力____的频率,与系统的固有频 率___无__关____。
摆回到最高点,所以周期为1.2 s,由T=2π l 可知该摆的摆长l= T2g =0.36 m,所
g
42
以选项D错误。
知识点二 共振的条件和规律 问题探究:
洗衣机在把衣服脱水完毕拔掉电源后,电动机还要转动一会才能停下来。在拔掉 电源后,发现洗衣机先振动得比较小,然后有一阵子振动得很剧烈,然后振动慢慢 减小直至停下来。 (1)开始时,洗衣机为什么振动比较小? (2)期间剧烈振动的原因是什么? 提示:(1)开始时,洗衣机的固有频率与脱水桶的频率相差较远。 (2)剧烈振动的原因是此时脱水桶的频率与洗衣机的固有频率接近。
机械振动学习题答案
2受迫振动
杆、轴、弦的受迫振动微分方程分别为
?2u?2u
杆：?a2?ea2?f(x,t)
?t?x?2??2?
轴：j2?gip2?f(x,t), j??ip
?t?x?2y?2y
弦：?2?t2?f(x,t)
?t?x
?n?1
(8)
(9)
下面以弦为例。令y(x,t)??yn(x)?n(t)，其中振型函数yn(x)满足式(2)和式(3)。代入式(9)得
lll
2
?n??n?n?
llqn(t)
, qn(t)??ynf(x,t)dx, b??yn2dx
00?b
(12)
当f(x,t)?f(x)ei?t简谐激励时，式(12)的稳态响应解为
qn(t)1l11i?t
?n(t)?yf(x)dxe?n2222?0?b?n???n???b全响应解为
?n(t)?
?1l1??
?d1sinkl1?c2coskl1?d2sinkl1
② ③
du1(l1)du2(l1)
?ea2 ?ad④ 11coskl1?a2?d2coskl1?c2sinkl1? dxdx
②式代入③式得d1tankl1?c2?1?tankl1tank(l1?l2)?
②式代入④式得所以频率方程即
d1?c2?tank(l1?l2)?tankl1?a2/a1
q(x)?ccoskx?
dsinkx，其中k?① ②
c?0, gipdkcoskl?t0 q(x)?
t0
sinkx
gipkcoskl
t0
sinkxsin?t
gipkcoskl
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u (t) (5cost 3)2 (5sin t)2 34 30cos t (t) arctan( 5sin t )
5cos t 3
umax 34 30 8 umin 34 30 2
拍频 | 2 1 || 40 39 | 1 rad/s 拍周期 2 2 2 (s)
g
g 9.8 9.9mm
mg
2 sin(t ) 2 (2 5)2
P57.1-3: 求简谐位移u1(t) 5e j(t300)与u2(t) 7e j(t900)的合成运动u(t), 并求u(t)与u1(t)的相位差。
u(t) u1(t) u2 (t) 5e j(t300 ) 7e j(t900 ) (5e j300 7e j900 )e jt (5 cos 300 j(5sin 300 7))e jt 10.44e j(t65.50 )
由以上各式得到：keq
(a b)2 a2 b2
k2 k1
k1x1 x1
a
bx1 ax2 ab
k 2 x2
o
x2
b f
P57.1-7: 图中简支梁长l 4m, 抗弯刚度EI 1.96106 Nm2, 且k 4.9105 N/m, m 400kg。 分别求图示两种系统的固有频率。
w
F F／2
/ m)
n
keq m
1.96 106 70(rad / s)
400
(b)
keq
k kbeam k kbeam
3.675105
n
keq 30.3(rad / s) m
P58.1-8: 钢索的刚度为4105 N/m, 绕过定滑轮吊着质量为100kg的物体以匀速0.5m/s下降， 若钢索突然卡住，求钢索内的最大张力。
对台面的振幅有何限制？
m
u
质量m运动方程：N mg mu&&(t)
N mu&&(t) mg
不跳离条件: N 0
a
sin(t
)
g
2
g u(t) 2
mu&&(t) mg 0
u&&(t) 2u(t)
(•) 如果sin(t ) 0, 则上式恒成立
N
m
(•) 如果sin(t ) 0, 则上式变为a
x3
12
1 6
x
l 2
3
3l 2 48
x
w
F F／2
F／2 x
w(x)
F EI
x3
12
1 6
x
l 2
3
3l 2 48
x
(a)
F
F 48EI
kbeam w(l / 2) l3F l3
48EI
keq
k
kbeam
k
48EI l3
4.9 105
48 1.96 106 43
1.96106 (N
F／2
x
任意截面处的弯矩：
M (x) F x F x l
2
2
挠曲线微分方程：
x
l 2
x 0
l 2
当x l 2
当x l 2
d 2w
M (x)Leabharlann F 2xFx
l 2
dx2 EI
EI
积分：
w(x)
F EI
x3
12
1 6
x
l 2
3
Cx
D
边界条件： w(0) w(l) 0
w(x)
F EI
| 2 1 | | 40 39 |
P57.1-5: 写出图示系统的等效刚度的表达式。当m 2.5kg, k1 k2 2105 N/m, k3 3105 N/m时， 求系统的固有频率。
分析表明：k1和k2并联，之后与k3串联 k1和k2并联后的等效刚度：keq k1 k2
整个系统的等效刚度：keq
振动周期：T 2 / 2 / 2.1167 2.9684 振幅：a 0.1069 最大速度=a 0.1069 2.1167 0.2263
[a2 0.052 ]2 0.22 [a2 0.12 ]2 0.082
解出
a 0.1069, =2.1167
P57.1-2: 一物体放在水平台面上，当台面沿铅垂方向作频率为5Hz的简谐振动时，要使物体不跳离平台，
系统固有频率：n
k m
初始条件：u(0) 0, u&(0) v0
振幅：a
u02
( u&0 )2 n
v0 n
v0
m k
最大张力：T mg ka mg v0 mk 1000 9.8 0.5 1000 4105 1.98104 (N)
P58.1-11: 系统在图示平面内作微摆动，不计刚杆质量，求其固有频率。
(ml2 2ml2 )&& k l2 mgl 4
n
kl 4mg 12ml
P58.1-12: 图示摆，其转轴与铅垂方向成角，摆长l，质量不计。求摆动固有频率。 ml2&& mg sin( )l sin
ml2&& mg sin( )l sin 0
很小，sin
ml2&& mg sin( )l 0
第一章习题
P57.1-1: 一物体作简谐振动, 当它通过距平衡位置为0.05m, 0.1m处时的速度分别为0.2m/s和0.08m/s。 求其振动周期、振幅和最大速度。
u(t) a sin(t ) u&(t) a cos(t )
两边平方，相加
代入已知条件
[a2 u2 (t)]2 u&2 (t)
n
mg sin( )l ml 2
g sin( ) l
mg sin
P58.1-13: 证明对临界阻尼或过阻尼，系统从任意初始条件开始运动至多越过平衡位置一次。
(1) 对临界阻尼情形
u(t) [u0 (u&0 nu0 )t]ent
keq k3 keq k3
(k1 k2 )k3 k1 k2 k3
系统的固有频率：n
keq 261.86 rad/s m
P57.1-6: 写出图示系统的等效刚度的表达式。
垂直方向力平衡：f k1x1 k2x2
对o力矩平衡：k1x1a k2x2b
设等效刚度系数为keq，则：f
keq
bx1 ax2 ab
u(t)与u1(t)的相位差：65.50 300 35.50
P57.1-4: 求两简谐运动u1(t) 5cos 40t, u2 (t) 3cos 39t的合成运动的最大振幅和最小振幅， 并求其拍频和周期。
u(t) u1(t) u2 (t) Re[5e j40t 3e j39t ] Re[(5e jt 3)e j39t ] Re[((5cos t 3) j5sin t)e j39t ] Re[u (t)e j (t)e j39t ]