Exercise of mechanical vibration (Vibration of 2-dof system)

Q1: Consider the following system consisting of two pendulums and aspring. Neglect the mass of rods and rotational friction of theirpivots. Answer the following problems.1) Answer the equations of motion about θ1 and θ2.2) Answer the natural frequencies of the system.3) Answer the mode shapes corresponding to the two natural

frequencies.

m m g

θ1 θ2

la k

Basic

Q2: Answer the general solutions of the following 2‐D.O.F about x1 andx2. model, which appears in page 84 of the text book, under twoinitial conditions.

1) 𝑥 0 𝑥 0 𝑎, 𝑥 0 𝑥 0 02) 𝑥 0 𝑎, 𝑥 0 𝑎, 𝑥 0 𝑥 0 0

𝑥𝑘 𝑘 𝑘

𝑥

Basic

Q3: Suppose the following 2‐D.O.F. cart model, which appears in page 86of the text book. 𝑥 𝑥 0 at the equilibrium position of thesystem.1) Answer the equation of motion about x1 and x2.2) Answer the natural frequencies of the system by solving the

characteristic roots.3) Answer the two modes (mode shapes or vectors and 

corresponding natural frequencies) of the system.𝑥𝑘 𝑘

𝑥

Basic

Q4: Consider a pulley with a rope which has two springs at the ends. Therope does not slip on the pulley and is tensioned by a mass of m.The mass and radius of the pulley are M and R, respectively. θ = x =0 at the equilibrium position of the system. (problem 4.2b in p. 99)1) Answer the equations of motion of the system.2) Answer the natural frequencies of the system by solving the

characteristic equations.

k k

m

MR

θ

x

Basic

Q5: Consider a rigid bar that is supported by two springs at its ends. Themass moment of inertia of the bar is I around G, which is the centerof mass of the bar. The bar translates along x, and rotates around Gat angle θ, which is small enough to allow us to use a small angleassumption. The effect of the gravity is ignored. (Example in p. 88)1) Answer the equations of motion of the system about x and θ.2) Answer the natural frequencies of the system.

k2k1

a b

θ

xx-aθ x+bθG

Intermediate

Q6: Consider a rod of length 2l with two mass points at the ends. Theposition of the center of the rod is denoted by x, which is 0 at theequilibrium position. The rotation angle of the rod is θ. (Q. 4.3a in p.98)1) Answer the equations of motion of the system.2) Answer the natural frequencies of the system when θ is small.

k k

ll θmm x

Intermediate

Q7: Objects of mass m are connected by strings of which tension is T.The length of the strings is l. The displacements of two objects arey1 and y2 and they are small enough that each string functions as alinear spring of constant T/l. (Q. 4.4a, p. 99)1) Answer the equation of motion of the system about y1 and y2.2) Answer the two natural frequencies of the system.3) Answer the ratio of amplitudes of y1 and y2 for each natural

frequency.

l mm lly1 y2

Basic

Q8: Suppose two identical frictionless mass m are connected by a springof constant k. The spring is at its free length when 𝑥 𝑥 .1) Answer the equation of the motion about 𝑥 and 𝑥 .2) Answer the natural frequency of the system.3) Answer the distance between two mass points, 𝑥 𝑥 , when𝑥 0 𝑥 0 0 and 𝑥 0 𝑥 0 0.

𝑥𝑘

𝑥

Basic

Q9: A mass is supported by two springs, one of which is angled at 𝜋/4.The displacement from the equilibrium position is denoted by x1and x2. The displacement is small enough that the angle of thespring is considered constant. Answer the following problem.1) Answer the equation of the motion about 𝑥 and 𝑥 .2) Answer the natural frequencies of the system.3) Answer the modal shapes or vectors corresponding to the

natural frequencies found at 2). 𝑥𝑥

Basic

𝑘𝑘 𝜋/4 m 𝑥𝑘 𝑐

m𝑘 𝑥

Q11: Consider three objects coupled bysprings and a damper as shown in thefigure. The vertical motion of the objectsare expressed by 𝑥 , 𝑥 and 𝑥 . When thesystem is at the equilibrium position, 𝑥𝑥 𝑥 0 . Answer the equations ofmotion about 𝑥 , 𝑥 , and 𝑥 .

m𝑘 𝑥

Basic

Q13: Consider the following 2‐DOF system consisting of m1 and m2. Twoharmonic forces f1 and f2 act on each mass. Consider only the specialsolution.1) Answer the amplitude of the force transmitted to the base FT.2) Answer the amplitude of FT when 𝜔 𝑘 /𝑚 .

m1

𝑓 𝐹 cos 𝜔𝑡m2𝑓 𝐹 cos 𝜔𝑡 𝑥

𝑥𝐹 𝑘𝑘

Intermediate

Q14: Consider a mass‐spring system shown in the figure. The mass isvibrated by a harmonic displacement through a spring. m = 10 kg, k1= 3,000 N/m, k2 = 6,000 N/m, and c = 300 N・s/m. The harmonicdisplacement is 𝑦 acos 𝜔𝑡 where 𝑎 0.002 m and 𝜔 8𝜋 rad/s.1) Answer the equation of motion of the system.2) Answer the motion of the mass.3) Answer the reaction force received at P.

m

𝑦 𝑎cos 𝜔𝑡𝑥𝑘

𝑘 P𝑐

Intermediate

Hint: Solution of the following equation of motion is given by𝑥 𝑡 𝑋cos 𝜔𝑡 𝛽𝑋 𝛿1 𝑟 2𝜁𝑟𝛽 𝑇𝑎𝑛 2𝜁𝑟1 𝑟𝑟 𝜔𝑝𝛿 𝐴𝑘

𝑚𝑥 𝑐𝑥 𝑘𝑥 𝐴cos𝜔𝑡.

Q15: Consider a rigid bar that is supported by two springs at its ends.This system is regarded as a 2‐dof system. The bar translates along x,and rotates around G (center of the mass) at angle θ, which is smallenough to allow us to use a small angle assumption. The effect ofthe gravity is ignored.1) Find the equations of motion of the system about x and θ.2) Find the natural frequencies of the system.3) Find the modal vectors of the system.

2kk

a a

θ

xx-aθ x+aθG

Intermediate

4) When the harmonic force (𝐹cos𝜔𝑡) is applied to the bar alongthe x axis, find the amplitudes of 𝑥 and 𝜃.

5) Find the node of the bar when 𝜔 .

2kk

a a

θ

xx-aθ x+aθG

Intermediate

Q16 A cart (m) is supported by a spring (k) on another cart (M). Motionsof the carts are restricted on the horizontal direction. The positionof M in the reference coordinate system is denoted by 𝑥. Thedeflection of the spring is 𝑦, and 𝑦 0 when the spring is of itsnatural length.(1) Find the equation of motion about 𝑥 and 𝑦.(2) Find the natural frequency 𝜔 of the system.(3) Find the mode vector at 𝜔 .(4) Solve 𝑥 and 𝑦 with the initial conditions: 𝑥 0 0, 𝑥 0 0,𝑦 0 𝑦 , and 𝑦 0 0.

(Deflection of the spring)

Q17: The figure shows a vibration absorber that is used on somepassenger cars to reduce the vibration of the exhaust pipe. Supposethat the forcing frequency is 5 Hz. If we use an absorber mass 𝑚0.3 kg, what must be the value of the stiffness of the cantileverbeam.

Exhaust pipe

m2

Cantilever

Dampermass

https://www.physicsforums.com/

PipeMass

Basic

Q18: Consider a frictionless cart that is supported by springs and adamper. Their spring and damping coefficients are k, c1, and c,respectively. The displacement of each point is 0 when the system isstill with no external forces. Answer the following problems.

1) Given that u1 and x2 are fixed to zero, answer the equation ofmotion of the cart.

𝑥𝑘

𝑘𝑥𝑢

Intermediate

2) Answer the characteristic roots of the equation acquired in 1), aswell as the damped natural frequency 𝜔 and damping ratio 𝜁which is smaller than 1.

3) When u and x2 are free (not fixed to zero) and c1 = 0, answer theequation of motion.

4) Point u is forced to vibrate at frequency 𝜔: 𝑢 𝐴sin 𝜔𝑡 .Answer the vibratory amplitude of the cart (amplitude of x1).

Q19: Two rotary disks, which rotate around their center of gravity, are connected by a string of spring coefficient k. The rotary angles are denoted by 𝜃 and 𝜃 , of which angles are zero at their equilibrium positions. The mass moment of inertia of disks 1 and 2 are 𝐼 and 𝐼 , respectively. Answer the following problems.

𝜃 𝜃kr r

Disk 1 Disk 2

Basic

(1) Answer the equation of motions of the system about 𝜃 and 𝜃 .(2) Answer the natural frequencies of the system: 𝜔 and 𝜔 .(3) One of the natural frequencies found in (2) is 𝜔 0. Answer the 

mode vector for 𝜔 . This mode is referred to a rigid‐body mode. Describe how the disks behave for this mode.

(4) Answer the mode vector for 𝜔 .

Q21: Answer the problems about the following 2‐D.O.F system about x1and x2. We only consider the special solutions.1) Find the excitation frequency 𝜔 with which 𝑥 0 when 𝑓𝑓cos𝜔 𝑡.2) Answer 𝑥 under the condition of problem 1).

𝑥𝑘 𝑘 𝑘

𝑥𝑓

Basic

Q22 Consider a three‐cart system. Answer the natural frequencies andmode vectors of the following system (Example 5.1 in the text book).

𝑥𝑘 𝑘

𝑥𝑘

𝑥𝑘

Intermediate

Q23 A homogeneous disk (mass moment of inertia 𝐽 ) is supported by arod of which torsional stiffness is 𝑘 . An excitation torque 𝑇𝜏sin𝜔𝑡 is applied to the disk, and its rotation angle is 𝜃 .(1) Find the natural frequency 𝜔 of the system.(2) Find the excitation frequency 𝜔 (𝜔 𝜔 ) at which the

amplitude of 𝜃 is 2𝜏/𝑘 .We attach an additional disk (moment of inertia: 𝐽 ) and rod(torsional stiffness: 𝑘 ) in order to achieve a dynamic absorber forthis system. The torsion of the additional rod is 𝜃 . 𝜃 0 when thesystem is still with no excitation torque applied. Answer thefollowing problems.(3) Find the equation of motion about 𝜃 and 𝜃 .(4) Find the relationship of 𝑘 , 𝐽 and 𝜔 for which 𝜃 0.(5) Under the condition of (4), find the amplitude of 𝜃 .

Basic

𝑇 𝑡 𝜏sin𝜔𝑡𝑘 𝐽𝜃

𝑇 𝑡 𝜏sin𝜔𝑡𝑘 𝐽𝜃

𝑘 𝐽𝜃

Q24 A rotary engine is laid on a beam of which mass can be neglected.When the still engine is placed on the beam, the center of the beamdeflected 𝛿. The total mass of the engine is 𝑀. The mass of therotary part is 𝑚 and its moment arm is 𝑟. The angular velocity isconstant i.e. 𝜔. In order to mitigate the vibration of the engine, adynamic vibration absorber made of a couple of a spring and massis installed beneath the engine. The vertical displacement of thecenter of the engine and mass are 𝑥 and 𝑥 , respectively. Thesedisplacements are zero when the system is statically equivalent.(1) Find the spring coefficient of the beam when it is regarded as a

linear spring along the vertical direction.(2) Find the equations of motion about 𝑥 and 𝑥 .(3) Find the amplitude of the engine by solving the special solution

of the equations of motion.(4) Determine the relationship between 𝑚 and 𝑘 with which the

amplitude of the engine is most effectively reduced.Basic

m

Q25 We calculate the principal modes of a double pendulum system.The rotation angles of the masses are 𝜃 and 𝜃 , which are zerowhen the pendulum is still. Masses are 𝑚 and 𝑚 . The origin ofthe coordinate is laid on the pivot of the pendulum. Use g as thegravitational acceleration.(1) Express the coordinate (𝑥 and 𝑦 ) of 𝑚 using 𝑙 , 𝑙 , 𝜃 , and𝜃 .(2) Express the velocity of 𝑚 and 𝑚 using 𝑙 , 𝑙 , 𝜃 , and 𝜃 . Note,

the velocity of the i‐th mass is 𝑥 𝑦 / .(3) Find the equations of motion about 𝜃 , and 𝜃 by using

Lagrange’s method. Suppose that 𝜃 and 𝜃 are small and weapproximate trigonometric functions such that they do notinclude 2nd and higher‐order components.

(4) Find the natural frequencies and corresponding mode shapeswhen 𝑚 𝑚 𝑚 and 𝑙 𝑙 𝑙.

Intermediate

m1g

θ1

θ2

𝑙

m2

𝑙

Q26 Consider a pulley, around which a rope wound. The rope has twosprings at its ends. The rope does not slip on the pulley and istensioned by a mass of m. The mass and radius of the pulley are 3𝑚and R, respectively. θ = x = 0 at the equilibrium position of thesystem.1) Find the equations of motion of the system. Note that the mass

moment of inertia of the pulley is .2) Find the natural frequencies of the system by solving the

characteristic equations.3) Find the ratio of the amplitudes of 𝜃 to 𝑥 𝜃/𝑥 for each of the

two natural frequencies.4) Slowly pull m downward with the constant force f, and then

release it from rest. Find the motions of 𝜃 and 𝑥 as functions oft.

kk

m

3m

Rθ

x

S1 A pulley of m1 is supported by a spring of k1 from the ceiling as in thefigure. The radius of the pulley is R. A wire is wound by the pulley.One end of the wire is fixed to a spring (k2). Another end supports amass m2. The displacements of the pulley and mass, respectively,from their equilibrium positions are denoted by x1 and x2. Therotation angle of the pulley is 𝜃, and 𝜃 0 at the equilibriumposition.(1) Find the mass moment of inertia of the pulley.(2) Express x2 using x1 and 𝜃.(3) Express l, which is the deflection of spring 2, using x1 and 𝜃.(4) T1 and T2 are the tensions of the wire caused by the mass and

spring, respectively. Find the equations of motion of the pulleyabout x1 and 𝜃 using T1 and T2.

(5) Find the equation of motion about x2 using T1.(6) Find the relationship between l and T2.

大学院入試問題に挑戦しよう. 名古屋大学

(7) Find the equations of motion about x1 and x2 without using T1,T2 and 𝜃.

(8) Find the natural frequencies of the system when 𝑘 𝑘 𝑘and 𝑚 𝑚 𝑚 .

大学院入試問題に挑戦しよう. 名古屋大学

R

k1

x1m1

k2 m2 x2

R

k1

x1m1

S2 Two disks are connected by a string and springs of coefficient 𝑘. Eachdisk rotates around 𝑂 and 𝑂 . The mass moments of inertia are 𝐼and 𝐼 . The string is not loose and properly tensioned. The string iswound around the disks and does not slip. The rotation angles oftwo disks are denoted by 𝜃 and 𝜃 which are zero at theequilibrium state of the system.(1) Find the equations of motion about 𝜃 and 𝜃 .(2) Find the natural frequencies and corresponding mode vectors of

the system.

大学院入試問題を解いてみよう. 2017年 九州大学大学院入試より

𝜃 𝜃k

k

𝑅𝑅 𝑂 𝑂

S3 A cart (m1) and circular disk (m2) are connected by springs as in thefigure. The radius and mass of the disk are r and m1, respectively.The disk is supported by the springs at its center in a frictionlessmanner. The position of the cart and rotation angle of the disk aredenoted by x and θ, respectively. They are zero at the equilibriumstate of the system. The disk does not slip. Answer the followingproblems.(1) Find the mass moment of inertia I of the tire.(2) Find the kinetic (T) and potential (U) energies of the system.(3) Find the equations of motion of the system.(4) Find the natural angular frequencies of the system using m2 and

k2 when 𝑚 𝑚 and 𝑘 3𝑘 .(5) Find the ratios at the natural angular frequencies found in (4).

大学院入試問題を解いてみよう. 2016年度 東北大学 大学院入試問題

𝜃𝑘 𝑘

𝑥

S4 Wilberforce pendulum consists of a mass and a long coil spring,which extends along its longitudinal axis while unwinding. Incontrast, the shrink of the spring causes it to wind tighter. Thesetwo types of motions, which are translational and torsional motions,are coupled and the kinetic energy shifts between them. Thevertical displacement of the mass is denoted by x, and its torsionangle is by 𝜃. The weight and inertia of the mass are m and I,respectively. The equations of motion of the system are expressedby

where 𝑘 and 𝑘 are the spring coefficient for extension androtation, respectively. Also, 𝑘 is a constant to couple the torsionand vertical displacement. x and 𝜃 are zero at their equilibriumpositions.

𝑚𝑥 𝑘 𝑥 𝑘 𝜃 0𝐼𝜃 𝑘 𝑥 𝑘 𝜃 0

大学院入試問題を解いてみよう．2017年度 名古屋大学

When, 𝜔 , answer the following questions.

1) Answer the characteristic equation of the system.

2) Answer all the natural frequencies of the system using m, I, kd,and 𝜔 .

3) Because two natural frequencies of the system are close, a beatphenomenon occurs. The vertical and torsional motionsalternate at a low frequency. Answer the beat frequency 𝜔 .

4) Explain the reason that 𝜔 ≪ 𝜔 when 𝑘 ≪ 𝑘 𝑘 .Hint: 1 𝑥 ∼ 1

* Beat （うなり）という現象は，講義では時間の都合で割愛したが，よく知られた現象であるので，テキストを参照して，勉強してください．

You are recommended to see how Wilberforce pendulum behaves onthe Internet.

x

θ

S5  2つの物体が2つのばね（ばね定数2k, k）で図2のように連結された系を考える．物体1（質量2m）と物体2（質量m）が静的に釣り合った位置からのそれぞれの変位を𝑥 , 𝑥 とする．物体は摩擦の無い床面を滑る．1) 𝑥 , 𝑥 に関する運動方程式を示せ．2) 物体2に右向きに励振力 𝐹cos𝜔𝑡 が加わるとき（Fは定数），物体2の変位の振幅が0となるようなωをm, kを用いて表し，また，このときのx_1をtの関数で示せ．

3) 系の固有角周波数を小さい順から全て示せ．4) 問3)で示した固有角周波数に対応するモードベクトルを示せ．な

お，モードベクトルは (1, A) のように， 𝑥 の振幅が1になるような形で示せ．

5) 初期条件が 𝑥 , 𝑥 1,0 ,  𝑥 , 𝑥 0, 1 であるときの，𝑥 , 𝑥 の自由振動解（外力が作用しない時の解）を tの関数で示せ．

大学院入試問題を解いてみよう．2019年度 名古屋大学

𝑥2𝑘 𝑘

𝑥 𝐹cos𝜔𝑡A mode of an oscillating system is a pattern of motion in which allparts of the system move sinusoidally with the same frequency andwith a fixed phase relation. These frequencies of the modes areknown as its natural frequencies. A physical object, such as abuilding or bridge has a set of modes and their natural frequencies.A mode of vibration is characterized by a modal frequency and amode shape. It is numbered according to the number of half wavesin the vibration. For example, if a vibrating beam with both endspinned displayed a mode shape of half of a sine wave (one peak onthe vibrating beam), it is vibrating in the first mode. If it had a fullsine wave (one peak and one trough) it is vibrating in the secondmode.

QA: Translate the following paragraph about the mode of vibration.

sinusoidally … 正弦波状に（で）, mode shape … モード形状

A dynamic vibration absorber is useful for situations in which thedisturbance has a constant frequency. As opposed to a vibrationisolator, which contains stiffness and damping elements, a vibrationabsorber is a device consisting of another mass and a stiffnesselement that are attached to the main mass to be protected fromvibration. If we know the frequency of the disturbing input and thenatural frequency of the original system, we can select the valuesfor the absorber’s mass and stiffness so that the motion of the mainmass is very small, which means that its kinetic and potentialenergies are small. The energy delivered by the disturbing inputmust be absorbed by the absorber. Thus the resulting absorbermotion will be large. Another term used for vibration absorber is amass damper.

QB: Translate the following paragraph about the mode of vibration. COLUMN: 学びの動機付け

教育学の研究論文によると，学習の動機付けは大きく２つに分けられます．１つ目は，内的動機（Intrinsic motivation）です．これは，学

習の楽しさに代表されます．知識を得る，理解が進む，問題が解けるようになるという実感が，学習意欲につながるということです．２つ目は，外的動機（Extrinsic motivation）です．これは，その学問が将

来の仕事や試験の役に立つ（利用価値），他の人より良い成績を修めることが重要であるということが，学習意欲になるということです．

私が学生の時，ある科目に全く興味が持てず，勉強もしませんでした．今思えば，その科目については，内的動機も外的動機も欠いていました．皆さんのそれぞれの科目に対する動機付けはどのようになっていますか？

• この演習問題には解答例に誤りが含まれている場合があります．初めに，誤りを見つけてくれた人には，最終成績に2点を加点します．2問目以降は1点ずつを加点します．

Answer:

Q11), 2, 3) テキスト p. 89‐90の4.2.3 c) の例題と同じであるので割愛する．

ヒント: 特性方程式は下記の通りである．𝑚 𝑙 𝜔 2𝑚𝑙 𝜔 𝑚𝑙𝑔 𝑘𝑎 2𝑚𝑔𝑙𝑘𝑎 𝑚 𝑔 𝑙 0

Q2

1)

2)

𝑥𝑥 𝑎𝑎 cos 𝑘𝑚 𝑡 一次モードのみが発現する

𝑥𝑥 𝑎𝑎 cos 𝑘 2𝑘𝑚 𝑡 二次モードのみが発現する

すなわち，初期条件によってモードの現れ方が変化する

ヒント: 次の一般解の未定数を初期条件によって決定する𝑥𝑥 𝐴 11 cos 𝑘𝑚 𝑡 𝜙 𝐴 11 cos 𝑘 2𝑘𝑚 𝑡 𝜃

Q3テキストのp. 86‐87 の例題とおりであるので，解答を割愛する．

Q41) 𝐼𝜃 𝑘𝜃𝑅 𝑘𝑅 𝑅𝜃 𝑥 0𝑚𝑥 𝑘 𝑥 𝑅𝜃 02) 𝑝 , 𝑝 𝑀𝑘 4𝑚𝑘 𝑘 𝑀 16𝑚2𝑀𝑚

k

m

MR

θ

x

このばねのたわみは，𝑥 𝑅𝜃であるので，mに関する運動方程式は，𝑚𝑥 𝑘 𝑥 𝑅𝜃 0

𝑓 𝑘𝑅𝜃 𝑓 𝑘 𝑥 𝑅𝜃𝑅 𝑓 𝑓 𝐼𝜃 プーリの両端からの張力の差が，プーリを回転

させるトルクになる

Q4解説つづき

Q4(1)

・並進運動

物体の並進運動とプーリの回転運動について運動方程式を立てる

𝑚𝑥 𝑘 𝑥 𝑅𝜃右のばねの伸びは𝑥 𝑅𝜃なので，

𝑚𝑥 𝑘 𝑥 𝑅𝜃 0・回転運動

左のばねによるモーメントは 𝑘𝑅𝜃 𝑅，右のばねによるモーメントは𝑘 𝑥 𝑅𝜃 𝑅なので，𝐼𝜃 𝑘𝑅𝜃 𝑅 𝑘 𝑥 𝑅𝜃 𝑅𝐼𝜃 𝑘𝑅 𝜃 𝑘𝑅 𝑥 𝑅𝜃 0これらより，𝑚 00 𝐼 𝑥𝜃 𝑘 𝑘𝑅𝑘𝑅 2𝑘𝑅 𝑥𝜃 00

k k

m

MR

θ

x

𝑓 𝑘𝑅𝜃 𝑓 𝑘 𝑥 𝑅𝜃Q4

(2) 系の固有角振動数を𝑝とし，解を𝑥𝜃 𝑋Θ cos 𝑝𝑡 𝜙 と仮定する

(1)で得られた運動方程式に代入すると，𝑝 𝑚 00 𝐼 𝑋Θ 𝑘 𝑘𝑅𝑘𝑅 2𝑘𝑅 𝑋Θ cos 𝑝𝑡 𝜙 00𝑘 𝑚𝑝 𝑘𝑅𝑘𝑅 2𝑘𝑅 𝐼𝑝 𝑋Θ 00

cos 𝑝𝑡 𝜙 の係数比較から，

係数マトリックスの行列式より，特性方程式が得られる𝑚𝐼𝑝 𝑘𝐼 2𝑚𝑘𝑅 𝑝 𝑘 𝑅 0ここで，𝐼 𝑀𝑅 を代入して変形すると，特性方程式は，

𝑝 𝑀𝑘 4𝑚𝑘 𝑘 𝑀 16𝑚2𝑀𝑚 , 𝑝 𝑀𝑘 4𝑚𝑘 𝑘 𝑀 16𝑚2𝑀𝑚𝑀𝑚𝑝 𝑘 𝑀 4𝑚 𝑝 2𝑘 0

となる．これより，求める固有角振動数は，

Q5テキストのp. 88 の例題とおりであるので，解答を割愛する．

Q6テキストのp. 98 の問題であるので，解答を割愛する．

Q7

テキストのp. 99の 4.4a のとおりであるので，解答を割愛する．1), 2)

3) 𝑝 𝑇𝑚𝑙𝑝 3𝑇𝑚𝑙

に対する，y1とy2の振幅比は，1:1.

に対する，y1とy2の振幅比は，1:‐1 （もしくは ‐1:1）．

Q81) 𝑚𝑥 𝑘𝑥 𝑘𝑥 0𝑚𝑥 𝑘𝑥 𝑘𝑥 02) 𝑝 2𝑘𝑚 固有周波数

は１つである

3) 𝑥 𝑥 0 cos 𝑝𝑡𝑥 𝑥 0 cos 𝑝𝑡𝑥 0 𝑥 0 より，𝑥 𝑥 2𝑥 0 cos 𝑝𝑡2物体間の距離が振動する．

Q9

1)

2)

3)

𝑚𝑥 𝑘𝑥 12 𝑘𝑥 12 𝑘𝑥 0𝑚𝑥 12 𝑘𝑥 12 𝑘𝑥 0𝜔 , 𝑘𝑚 1 22

𝑥𝑥 11 2 , 12 1

1st mode

2nd mode

Q11 𝑚𝑥 𝑐𝑥 2𝑘𝑥 𝑘𝑥 0 𝑚𝑥 2𝑘𝑥 𝑘 𝑥 𝑥 0 𝑚𝑥 𝑘𝑥 𝑘𝑥 0

Q13(1)

調和外力が作用するばね・質量系

𝑚 𝑥 𝑘 𝑥 𝑘 𝑥 𝑥 𝑓質量𝑚 , 𝑚 の物体について，それぞれ運動方程式は，

𝑚 𝑥 𝑘 𝑘 𝑥 𝑘 𝑥 𝐹 cos 𝜔𝑡𝑚 𝑥 𝑘 𝑥 𝑥 𝑓𝑚 𝑥 𝑘 𝑥 𝑘 𝑥 𝐹 cos 𝜔𝑡これらより，𝑚 00 𝑚 𝑥𝑥 𝑘 𝑘 𝑘𝑘 𝑘 𝑥𝑥 𝐹𝐹 cos 𝜔𝑡強制振動解を

𝑥𝑥 𝑋𝑋 cos 𝜔𝑡 とおき，運動方程式に代入すると，𝑘 𝑘 𝑚 𝜔 𝑘𝑘 𝑘 𝑚 𝜔 𝑋𝑋 cos 𝜔𝑡 𝐹𝐹 cos 𝜔𝑡𝑘 𝑘 𝑚 𝜔 𝑘𝑘 𝑘 𝑚 𝜔 𝑋𝑋 𝐹𝐹cos𝜔𝑡の係数比較から，

Q13(1)の続き

調和外力が作用するばね・質量系

𝑋𝑋 について解くと，𝑋𝑋 1𝐷 𝜔 𝑘 𝑚 𝜔 𝑘𝑘 𝑘 𝑘 𝑚 𝜔 𝐹𝐹ここで，𝐷 𝜔 𝑘 𝑚 𝜔 𝑘 𝑘 𝑚 𝜔 𝑘 とおいた

これより，𝑋 𝑘 𝑚 𝜔 𝐹 𝑘 𝐹𝐷 𝜔𝐹 𝑘 𝑋 𝑘 𝑘 𝑚 𝜔 𝐹 𝑘 𝑘 𝐹𝑘 𝑚 𝜔 𝑘 𝑘 𝑚 𝜔 𝑘(2) 𝜔 𝑘 /𝑚 とおくと，𝐹 𝑘 𝑋 𝑘 𝑘 𝐹𝑘 𝐹 𝑘𝑘 副系の固有振動数と外力の周波数

が一致するとき，基礎に伝わる力は，主系に加わる力 f1 と無関係になる

Q132) 𝐴 𝑘 𝐹 𝑘𝑘

副系の固有振動数と外力の周波数が一致するとき，基礎に伝わる力は，主系に加わる力 f1 と無関係になる．

Q141) 𝑚𝑥 𝑐𝑥 𝑘 𝑥 𝑘 𝑥 𝑦 0 もしくは𝑚𝑥 𝑐𝑥 𝑘 𝑘 𝑥 𝑎𝑘 cos 𝜔𝑡2) 𝑝 30 rad/s, 𝜁 0.5, 𝛿 10 m𝑥 1.5 10 cos 8𝜋𝑡 1.233) 𝐹 𝑘 𝑦 𝑥6000 2 10 cos 8𝜋𝑡 1.5 10 cos 8𝜋𝑡 1.2312.4 cos 8𝜋𝑡 0.76 N

Q15

テキスト99ページの演習問題であり，解答も用意されているので，詳細は割愛する．なお，解答中のDは，行列式の値である．

(5) xの振幅Xが正, θの振幅Bが負であるので，図のような振動の形状になる．バーの重心から節までの横方向の距離を zとすると，tan𝐵 𝑋𝑧

が成り立つ．Bは十分に小さいとして， 𝑧𝐵 𝑋 の関係が成立するので，これを用いて，zを解く．

XB

節

z

類似した問題が，2018年夏の大学院入試に出題されました．

Q16(1)

(2) 𝜔 𝑘 1𝑀 1𝑚𝑀𝑥 𝑚 𝑥 𝑦 0𝑚 𝑥 𝑦 𝑘𝑦 0or

𝑀𝑥 𝑘𝑦 0𝑚 𝑥 𝑦 𝑘𝑦 0(3) 𝑥𝑦 𝑚𝑀 𝑚1(4) 𝑥𝑦 𝐴 𝑚𝑀 𝑚1 cos 𝜔 𝑡 𝜙 𝐵0解の形を として

Q16(4) Cont.𝑥 𝑚𝑦𝑀 𝑚 1 cos𝜔 𝑡𝑦 𝑦 cos𝜔 𝑡

運動方程式には，𝑥の項しかないため，𝑥には定数項があることがうかがえる．また，カートMは，どこにも固定されていないため，

原点から移動してしまう．これを表現するためには定数項が必要である．

Q17 𝜔 5 ⋅ 2𝜋 rad/s . から，𝑘 296 N/m.

Q18

1) 𝑚𝑥 𝑐 𝑥 2𝑘𝑥 02) 𝜆 𝑐2𝑚 𝑐2𝑚 2𝑘𝑚 𝜁𝜔 𝑖𝜔 1 𝜁

𝜔 2𝑘𝑚 𝜁 𝑐2 2𝑚𝑘 𝜔 𝜔 1 𝜁3) 𝑚𝑥 2𝑘𝑥 𝑘𝑥 𝑘𝑢 𝑘𝐴sin𝜔𝑡𝑐𝑥 𝑘 𝑥 𝑥 0𝑚 00 0 𝒙 0 00 𝑐 𝒙 2𝑘 𝑘𝑘 𝑘 𝒙 𝑘𝐴sin𝜔𝑡0 𝒙 𝑥𝑥

Q18

4) 𝜔 𝑚 2𝑘 𝑘𝑘 𝑘 𝑖𝜔𝑐 𝐵𝐵 𝑒 𝑘𝐴0 𝑒 𝒙 𝐵𝐵 𝑒𝐵𝐵 1𝜔 𝑚 2𝑘 𝑘 𝑖𝜔𝑐 𝑘 𝑘 𝑖𝜔𝑐 𝑘𝑘 𝜔 𝑚 2𝑘 𝑘𝐴0𝐵 𝑘𝐴 𝑘 𝑖𝜔𝑐𝜔 𝑚 2𝑘 𝑘 𝑖𝜔𝑐 𝑘 𝑘𝐴 𝑘 𝑖𝜔𝑐𝑘 𝜔 𝑚𝑘 𝑖𝜔𝑐 𝜔 𝑚 2𝑘

𝑘𝐴 𝑘 𝑖𝜔𝑐R 𝑖I 𝑘𝐴 𝑘 𝑖𝜔𝑐 R 𝑖IR I|𝐵 | 𝑘𝐴| 𝑘R 𝜔𝑐I 𝑖 𝜔𝑐R 𝑘IR I | 𝑘𝐴 R I 𝑘 𝜔 𝑐R I

𝑘𝐴 𝑘 𝜔 𝑐𝑘 𝜔 𝑚𝑘 𝜔 𝑐 𝜔 𝑚 2𝑘

Q211)

2) 𝑥 𝑓𝑘 cos 2𝑘𝑚 𝑡𝜔 2𝑘𝑚

Q22The solutions are shown in p. 103 of the text book.

Q23(1) (2)𝜔 𝑘𝐽 𝜔 𝑘2𝐽(3) 𝐽 𝜃 𝑘 𝑘 𝜃 𝑘 𝜃 𝜏sin𝜔𝑡𝐽 𝜃 𝑘 𝜃 𝑘 𝜃 0(4) 𝜃𝜃 𝑎𝑎 sin𝜔𝑡 として(3)に代入すると，下記が得られる

𝑘 𝑘 𝐽 𝜔 𝑘𝑘 𝑘 𝐽 𝜔 𝑎𝑎 sin𝜔𝑡 𝜏0 sin𝜔𝑡

Q23(4) つづき𝑎𝑎 1D 𝑘 𝐽 𝜔 𝑘𝑘 𝑘 𝑘 𝐽 𝜔 𝜏0

D 𝑘 𝑘 𝐽 𝜔 𝑘𝑘 𝑘 𝐽 𝜔𝑎 1D 𝑘 𝐽 𝜔 𝜏 より，𝑘 𝐽 𝜔 0 のときに，𝑎 0 となり，𝜃 0.(5) 𝑎 𝜏𝑘 マイナスは無くても良い

Q24(1) 𝐾 𝑀𝑔𝛿(2) 𝑀𝑥 𝐾 𝑘 𝑥 𝑘𝑥 𝑚 𝑟𝜔 sin𝜔𝑡𝑚𝑥 𝑘𝑥 𝑘𝑥 0 * もちろんcos関

数でも良い

(3)それぞれの振幅を𝑎 , 𝑎 とすると，𝑎𝑎 𝑚 𝑟𝜔D 𝑘 𝑚𝜔𝑘𝐷 𝐾 𝑘 𝑀𝜔 𝑘 𝑚𝜔 𝑘

(4) 𝑘 𝑚𝜔 0

Q25(1) 𝑥 𝑙 sin𝜃 𝑙 sin𝜃(2)

𝑦 𝑙 cos𝜃 𝑙 cos𝜃𝑥 𝑙 𝜃 cos𝜃 𝑙 𝜃 cos𝜃𝑦 𝑙 𝜃 sin𝜃 𝑙 𝜃 sin𝜃𝑣 𝑙 𝜃 𝑙 𝜃 2𝑙 𝑙 𝜃 𝜃 cos 𝜃 𝜃

Q25(3) 𝑇 12 𝑚 𝑙 𝜃 12 𝑚 𝑣 𝑈 𝑙 𝑚 𝑔cos𝜃 𝑙 cos𝜃 𝑙 cos𝜃 𝑚 𝑔

位置エネルギーはどこを基点とするかで変わるが，運動方程式に影響しない．𝑙 𝜃 𝑚𝑚 𝑚 𝑙 𝜃 𝑔𝜃 0𝑙 𝜃 𝑙 𝜃 𝑔𝜃 0cos 𝜃 𝜃 1 sin𝜃 𝜃 sin𝜃 𝜃ただし，次の近似を用いて，2次以上の項は含めない．

Q25(4)

D 𝑙 𝜔 𝑔 𝑀𝑙 𝜔𝑙 𝜔 𝑙 𝜔 𝑔 0特性方程式は下記のようになる． 𝑀 𝑚𝑚 𝑚𝜔 2 2 𝑔𝑙 なので，固有周波数は

𝜔 , 2 2 𝑔𝑙 となり，対応するモード形状は12 ,  12 である．

Q261) 3𝑚2 𝑅 𝜃 𝑘𝜃𝑅 𝑘𝑅 𝑅𝜃 𝑥 0𝑚𝑥 𝑘 𝑥 𝑅𝜃 02) 𝜔 , 𝑘3𝑚 , 2𝑘𝑚3) 𝜃𝑥 23𝑅 𝜃𝑥 1𝑅1st mode: 2nd mode:

4) 𝜃 0 𝑓𝑘𝑅 , 𝑥 0 2𝑓𝑘 , 𝜃 0 𝑥 0 0 として初期値問題を解く．

𝜃𝑥 9𝑓5𝑘 23𝑅1 cos 𝑘3𝑚 𝑡 𝑓5𝑘 1𝑅1 cos 2𝑘𝑚 𝑡 or

𝜃𝑥 6𝑓5𝑘𝑅 13𝑅2 cos 𝑘3𝑚 𝑡 𝑓5𝑘𝑅 1𝑅 cos 2𝑘𝑚 𝑡ボーナス問題のため，部分点は無しとします．完全に正答した場合のみ，5点を加点．

S1(1) 12 𝑚 𝑅 (2) 𝑥 𝑥 𝑅𝜃(3) 𝑙 𝑅𝜃 𝑥 (4) 𝑚 𝑥 𝑘 𝑥 𝑇 𝑇

𝑇 𝑚 𝑥𝐼𝜃 𝑇 𝑇 𝑅

(5) (6) 𝑙𝑘 𝑇(7) 𝑚 𝑥 𝑚 𝑥 𝑘 𝑥 2𝑥 𝑥 𝑘 012 𝑚 𝑥 𝑥 𝑚 𝑥 2𝑥 𝑥 𝑘 0

(8) 𝑚 1 112 32 𝑥𝑥 𝑘 3 12 1 𝑥𝑥 00𝜔 , 7 414 𝑘𝑚

S21) 𝐼 00 𝐼 𝜃𝜃 2𝑘𝑅 2𝑘𝑅 𝑅2𝑘𝑅 𝑅 2𝑘𝑅 𝜃𝜃 002) 𝜔 0, 𝜔

𝑿 1 , 𝑿 1

S3 12 𝑚 𝑟(1)

(2) 𝑇 12 𝑚 𝑥 12 𝑚 𝑟𝜃 12 𝐼𝜃 12 𝑚 𝑥 34 𝑚 𝑟𝜃𝑈 12 𝑘 𝑟𝜃 12 𝑘 𝑥 𝑟𝜃

(3)

32 𝑚 𝑟 𝜃 𝑘 𝑟 𝜃 𝑘 𝑟 𝑥 𝑟𝜃 0𝑚 𝑥 𝑘 𝑥 𝑟𝜃 0(4) 𝜔 , 

(5) 𝜔 のとき， 2𝜔 のとき， 2

S4

1) 𝜆 𝑚 𝑘 𝑘𝑘 𝜆 𝐼 𝑘 0 𝑘 𝜆 𝑚 𝑘 𝜆 𝐼 𝑘 02) 𝜔 , 𝜔 𝑘𝑚𝐼

3) 𝜔 𝜔 𝑘𝑚𝐼 𝜔 𝑘𝑚𝐼4) 𝜔 𝜔 1 𝑘𝜔 𝑚𝐼 𝜔 1 𝑘𝜔 𝑚𝐼

𝑘𝜔 𝑚𝐼 1𝜔 𝑘𝑚𝐼 ≪ 1𝜔 𝑘 𝑘𝑚𝐼 1𝜔 𝜔 𝜔∼ 𝜔 1 𝑘2𝜔 𝑚𝐼 𝜔 1 𝑘2𝜔 𝑚𝐼

𝜔 ≪ 𝜔

S5

2𝑚 00 𝑚 𝑥𝑥 3𝑘 𝑘𝑘 𝑘 𝑥𝑥 001)

𝑥 𝐹𝑘 cos 3𝑘2𝑚 𝑡 𝐹𝑘 cos 62 𝑘𝑚 𝑡2)

𝜔 𝑘2𝑚 , 2𝑘𝑚3)

𝑎𝑎 12 𝑎𝑎 114)

1次モード: 2次モード:

𝑥𝑥 𝐴cos𝜔 𝑡 𝐵sin𝜔 𝑡 12 𝐶cos𝜔 𝑡 𝐷sin𝜔 𝑡 11S55) 自由振動の一般解は

𝑥𝑥 13 cos 𝑘2𝑚 𝑡 13 2𝑚𝑘 sin 𝑘2𝑚 𝑡 12 23 cos 2𝑘𝑚 𝑡 13 𝑚2𝑘 sin 2𝑘𝑚 𝑡 11である．A-Dは初期値によって決定する値であり，解は

𝑥𝑥 13 cos 12 2𝑘𝑚 𝑡 13 2𝑚𝑘 sin 12 2𝑘𝑚 𝑡 12 23 cos 2𝑘𝑚 𝑡 16 2𝑚𝑘 sin 2𝑘𝑚 𝑡 11となる．有理化されていれば，下記の通りである．

QA振動系のモードとは，システムのすべての部分が同一の周波数と固定された位相関係で正弦波状に動くという運動のパターンである．これらのモード周波数は，システムの固有周波数として知られる．建物や橋梁などの物理的な物体は複数のモードと固有振動数を有する．振動のモードは，モード周波数と，モード形状によって特徴づけられる．モードは，振動の中の半波の数によって数字がつけられる．例えば，もし両端が固定された状態で振動している梁が半波形状（振動中の梁にピークが一つある）を示していれば，梁は一次モードで振動している．もし，振動中の梁が完全な正弦波の形状（一つのピークと谷）をしていれば，梁は二次モードで振動している．（原文: Wikipedia）

First mode of a vibrating beam Second mode