Exercise of mechanical vibration (Forced vibration)

𝑀

Q1 A frictionless cart is connected to the wall by way of a spring anddamper as is shown in the figure. On the cart of mass 𝑀, aneccentric motor is installed. The motor rotates at ω rad/s with aweight of 𝑚. The weight is supported by a weightless rod of length𝑙. 𝜃 is zero at 𝑡 0.

a) Answer the force generated by the motor along 𝑥 axis.

b) Answer the equation of motion of the cart about 𝑥.𝑚𝑘 𝑐𝑙

𝜃 𝜔𝑥

Basic

Q2  Obtain the equation of motion for each system about 𝑥 with 𝑥being a forced displacement. The equilibrium positon corresponds to 𝑥 𝑥 0.a) b)

Basic

Q3  Obtain the equation motion for each system about 𝜃. The equilibrium position corresponds to 𝜃 𝜃 0.a) b)

Neglect the inertia of the gears. The numbers of gear teeth are 𝑛 and 𝑛 .

A torsional spring

A torsional damper

Basic

Q4 Determine the equation of motion of the system shown in the figurewith the force 𝑓 as a given function of time. Assume that thedisplacement is small. The inertia of the rod is 𝐼.

𝑘𝑙 𝑙 𝜃 𝑐𝑓

Basic

Q5 Derive the equation of motion for 𝜃. The mass 𝑚 is attached to arigid rod having negligible mass and negligible pivot friction. Thedisplacement 𝑥 is a given function of time. When 𝑥 and 𝜃 are zero,the spring is at its natural length. Assume that 𝜃 is small.

𝑔𝑘

𝑙 𝑙𝑙𝑐

𝜃𝑥

Basic

Q6  Answer the following problems about the equation of motion:𝑚𝑥 𝑘𝑥 𝐹 sin 𝜔𝑡.a) Derive the solution when 𝑥 0 𝑥 0 0.b)   Derive the solution when 𝑥 0 0, 𝑥 0 𝑣 .

For the following questions, suppose the case where the system includes a small damping effect and we only consider the special solutions or steady state responses.c)   Answer 𝜔 with which the amplitude of the object is 2𝛿, where δ 𝐹 𝑘⁄ , the static deflection of the spring.d)   Answer 𝜔 with which the amplitude of the object is  𝛿.

Basic

Q7 Find the mass of 𝑚 when the system resonates.𝑘 3 10 N/m, 𝑘 1 10 N/m.𝑓 50 sin 20𝑡 N.𝑥

𝑘 𝑘Basic

𝑥𝑘

Q8  Find the range of angular velocity of 𝜔 when the amplitude of vibration is smaller than 2 mm. Suppose that the system has a small friction and the term of free vibration is negligible. In other words, we only consider the steady state of the system which is independent from the initial condition.𝑚 100 kg, 𝑘 3 10 N/m, 𝐹 500 N.

Basic

Q10 Consider a cart (𝑚 ) with another cart (𝑚 ) on it. They aresupported by springs as shown in the figure. 𝑥 and 𝑥 are theposition of two carts and 𝑥 , 0 at the equilibrium state. Answerthe equation of motion about each of 𝑚 and 𝑚 .

𝑥𝑘 𝑘

𝑥

Basic

Q11 Complete the rough sketch of the amplitudes and phases of theforced vibration of 1‐D.O.F. system.

When the equation of motion of an undamped 1‐D.O.F. system isgiven by 𝑚𝑥 𝑘𝑥 acos 𝜔𝑡,its steady state response becomes𝑥 𝛿1 𝑟 cos 𝜔𝑡where 𝛿 𝑎/𝑘 (static deflection) and 𝑟 𝜔/𝑝. p is the naturalangular frequency of the system. Draw the amplitude and phasecurves (or lines) of x using the following form.

Note that the sketched curves do not need to be precise; however,consider what kind of characteristics should be captured.

Basic

Amplitu

de

0

δ

𝑟 𝜔𝑝1

0

-π/2

-π

Phase

2

Q12 When the forcing frequency is 20 rad/s, a certain mass‐spring system oscillates with an amplitude equal to 50% of the amplitude if the forcing frequency is 10 rad/s. Estimate the natural frequency of the system.

Basic

𝑘

Q13  Consider the steady state response of the following 1‐D.O.F. system with harmonic force applied. When ζ = 1/4, answer the range of ω/p at which the amplitude of the cart is smaller than 2𝛿. 𝛿 𝑎/𝑘 . p is the natural angular frequency of the system, i.e. 𝑝 𝑘/𝑚.

𝑐Basic

Q14 Answer the problems about a cantilever which is composed of aweightless rod and a point mass m. It rotates around O with nofriction. The lever is supported by a couple of a spring (k) and adashpot (c). At the end of the lever, a harmonic force of frequency𝜔 is applied. The angle of the lever is denoted by 𝜃, and 𝜃 0 atthe equilibrium position.(1) Find the equation of motion about 𝜃.(2) Find the special solution of the equation acquired in (1).(3) Find the minimum value of damping coefficient c with which

the system does not resonate for any 𝜔 values.

Basic

𝑙

𝜃 m

𝑙𝑙𝐹cos𝜔𝑡

𝑘 𝑐𝑂

Q15 Consider the following cart with a spring and damper. The cart is excited by a harmonic displacement through the damper.1) Answer the equation of motion of the cart about x.2) Answer the vibratory amplitude of the cart as a function of ω.3) Answer the angular frequency when the amplitude becomes

maximum.4) Describe how c value influences the maximum amplitude.

(3.3 a in p. 77 of the text book)

𝑥 𝑎cos 𝜔𝑡Basic

Q16 A cantilever (片持ち梁) of mass m is excited by a harmonic force offrequency ω. The lever rotates around O with no friction. 𝜃 0 atits equilibrium position. The rotation angle is small enough thatsin 𝜃 ∼ 𝜃.1) Answer the equation of motion about θ.2) Answer the amplitude of the lever when it is at the steady state 

when l = 1 m, m = 2 kg, F0 = 100 N, k = 2×104 N/m, c = 30 N/m, and ω = 100 rad/s.

(3.2 b in p. 78 of the text book)

𝑙/2 𝜃k c

𝑙/2 𝐹 cos 𝜔𝑡Basic

Q17 The mass m is attached to the frame with a spring of stiffness k.The frame oscillates vertically with an amplitude of a at afrequency of ω.1) Answer the equation of motion about x.2) Compute the steady‐state amplitude of the mass when m = 0.5kg, k = 500 N/m, a = 4 mm, and ω = 6π rad/s.

𝑦 𝑎cos 𝜔𝑡 x

Basic

Q18 The equation of motion of a certain mass‐spring‐damper system is5𝑥 𝑐𝑥 10𝑥 𝑓 𝑡 .Suppose that f(t) = Fsinωt. We define M = X/δ, 𝛿 𝐹/𝑘 .1) Determine the natural frequency 𝜔 .2) Determine the resonance frequency 𝜔 and peak magnification

ratio Mr for ζ = 0.1 and 0.3.3) How large must the damping coefficient be so that the

maximum steady‐state amplitude of x is not greater than 3, ifthe input is f(t) = 22sinωt, for an arbitrary value of ω?

Basic

Q19 A certain mass‐spring system oscillates with an amplitude of 5 mmwhen the forcing frequency is 20 Hz, and with an amplitude of 1mm when the forcing frequency is 40 Hz. Estimate the naturalfrequency of the system. Suppose that the resonance frequency ofthe system is greater than 40 Hz.

Basic

Q20 Design an undamped isolator (絶縁器) for a 20‐kg mass subjectedto a harmonic forcing function whose frequency is 17 Hz. Theisolator should transmit to the base no more than 10% of theapplied force.

20 kg

𝑓 𝐹cos 𝜔𝑡1) Determine the maximum spring

coefficient.2) In practice, the vibration isolator

needs a damping element toreduce the vibration around theresonance point. Determine thedamping coefficient such that thetransmissibility is smaller than 3when p = ω. Use the springcoefficient acquired by 1).

Basic

Q21 Consider a machinery, of which mass is 120 kg, supported by aspring of coefficient 760 kN/m. The machine produces theharmonic force of 360 N at 3,600 rpm due to its imbalancedrotation. The damping ratio of this system is ζ = 0.18.1) Answer the amplitude of the vibration of the machinery.2) Answer the force transmissibility to the supporting base.3) Answer the amplitude of the force transmitted to the base.

Basic

Q22 Consider that a machine of mass m is supported by a spring ofcoefficient k. When the supporting base vibrates and itsdisplacement is Asinωt. Answer k when the amplitude of themachine becomes A/2.

m

𝐴sin 𝜔𝑡Basic

Q23 Consider a quarter model of automobile of which suspension iscomposed of a spring and damping element. The mass of the car ism. The car traverses a bumpy road whose surface is approximatedby a sinusoidal function of surface wavelength λ. Suppose that thecar resonates when the excitation frequency equals the naturalfrequency of the car. The wheel keeps its contact with the road.Answer the speed of the car v at which the car system getsresonated.

m

𝜆a

v

kc

Basic

Q24 Suppose that a car of mass m is supported by a shock absorber ofwhich stiffness and damping coefficient are k and c, respectively.The car runs on a sinusoidal surface at velocity v. The surfacewavelength of the road is λ. The position of the car from theaverage surface level is denoted by x. Answer the followingproblems.(1) Establish the equation of motion of the system about x.(2) Answer the natural frequency of the system given c = 0.(3) When 𝑐 0, answer the vibratory amplitude of the car at the 

steady state.

m

𝜆2δ

v

kcx(t)

Basic

Q25 Answer the steady‐state response of the system described by𝑚𝑥 𝑐𝑥 𝑘𝑥 𝑓 𝑐𝑜𝑠𝜔 𝑡 𝑓 sin 𝜔 𝑡where 𝑘 10 N/m, ζ = 0.1, 𝑟 𝜔 /𝑝 1/2, 𝑟 𝜔 /𝑝 3/2,and f1 = f2 = 1 N.

Basic

Q26 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 by P.

m

𝑦 𝑎cos 𝜔𝑡𝑥𝑘

𝑘 P

𝑐Basic

When the equation of motion of a damped 1‐D.O.F. system isgiven by 𝑚𝑥 𝑐𝑥 𝑘𝑥 acos 𝜔𝑡,its steady state response becomes𝑥 𝛿1 𝑟 2𝜁𝑟 cos 𝜔𝑡 𝛽𝛽 Tan 𝑐𝜔𝑘 𝑚𝜔where 𝜁 is the damping ratio of the system: . Draw theamplitude and phase curves using the following form, providedthat 0 𝜁 1/ 2.Note that the sketched curves do not need to be precise; however,consider what kind of characteristics should be captured.

Q27 Complete the rough sketch of the amplitudes and phases of theforced vibration of 1‐D.O.F. system.

Basic

Amplitu

de

0

δ

𝑟 𝜔𝑝10

-π/2

-π

Phase

2

R1 The figure shows a pendulum with a movable pivot. The weightlesspivot freely moves along the x‐axis with no friction. The length ofthe rod is l and its weight is negligible. θ = 0 at its equilibriumposition. The equation of motion for this system is obtained asfollow. The moment by the gravity and moving pivot is𝑀 𝑚𝑔𝑙sin𝜃 𝑚𝑥 𝑙cos𝜃.From the law of motion, 𝑀 𝑚𝑙 𝜃.Using the small‐angle‐assumption, the equation is rewritten as𝑚𝑙 𝜃 𝑚𝑔𝑙𝜃 𝑚𝑥 𝑙.By removing ml, we obtain𝑙𝜃 𝑔𝜃 𝑥 .Using the relation between θ and x,𝜃 𝑥 𝑥𝑙 ,

Intermediate

By removing θ and using x, we obtain 𝑥 𝑥 𝑥 .When the moving pivot is excitedas 𝑥 𝑎sin𝜔𝑡 , answer thefollowing problems.1) Answer the amplitude of this

system X using the naturalangular frequency p, ω, l, andg.

2) When 𝑥 𝑋sin𝜔𝑡 , themiddle point of the rod doesnot move. Find ω for thiscondition.

m

x0

O

x

θ

g

l

S1 Consider a vibration system shown in the figure. A point mass issupported by two wheels through pairs of a spring and dashpot.The mass vibrates along x‐axis. The wheels run on the wavysurfaces of spatial periods 𝑙 and 𝑙 . At 𝑡 0 , the wheels pass bypoints A and B at which the surfaces are highest. The position ofthe mass is denoted by x, and 𝑥 0 at its static equilibrium onpoints A and B. We consider the steady state of vibration. Answerthe following problems.

(1) Find the displacement of wheel 1 along x‐axis as a function of t.(2) Find the equation of motion of the system about x.(3) Find the velocity v at which the amplitude of the mass becomes

maximum when 𝑘 𝑘 𝑘, 𝑐 𝑐 𝑐, 𝑎 𝑎 𝑎, and𝑙 𝑙 𝑙.大学院入試問題を解いてみよう．2016年度東北大学大学院

m

𝑙a1

v

k1c1

A

B

𝑡 0

𝑙k2c2x

S2 Consider a rigid rod of zero weight rotating around pivot O. Themass m is attached at each end. The rotation angle of the rod isdenoted by θ, and θ = 0 at the equilibrium position. As is shown inthe figure, one end is connected to the wall by way of a spring‐damper unit. A forced displacement is applied to the other endthrough a damper. You may use a small angle assumption: sin 𝜃 ~𝜃.The gravity force does not act on the system. Answer the followingproblems.(1) Answer the mass moment of inertia of the rod about O.(2) Establish the equation of motion of the system about θ.(3) Answer θ as a function of t at the steady state.(4) At the steady state, find a condition under which the forced

displacement and the motion of mass 2 are synchronized withno phase difference.

大学院入試問題を解いてみよう．2016年度名古屋大学

k

c

m

m

O

θ

l

l

𝑥 𝑡 𝑋sin𝜔𝑡tc

mass 2

mass 1

大学院入試問題を解いてみよう．2016年度名古屋大学

A Translate the following paragraph about forced vibration intoJapanese.

Mechanical vibrations can transmit for long distances through thestructure of a building. A related problem is the isolation ofvibration‐sensitive machines from the normally occurringdisturbances in a building. Examples of sensitive machines includesurgical microscopes, electronic equipment, lasers, MRI units, andcomputer disk drives. For minimum transmissibility (maximumisolation), the excitation frequency should be as high above thenatural frequency of the isolation system as possible. The inclusionof damping has the greatest effect in the vicinity of resonance,decreasing the vibration amplitude. A curious effect of damping isthat it results in increased amplitude at frequencies greater than2𝑝.

B Translate the following sentences about resonance into Japanese.

When vibrations of an undamped system are initiated, the motion issustained at the system’s natural frequency without additional energyinput. Thus, when the frequency of excitation is the same as thenatural frequency, the total energy increases because of the workdone by the external force and leads to a continual increase inamplitude. When the frequency of excitation is different from thenatural frequency, the work input is necessary to sustain motion atthe excitation frequency.

Modified from: G. Kelly, Mechanical vibrations, Cengage Learning

C  Translate the following paragraph about the resonance of undamped system.

When the system is undamped and the frequency of the excitationcoincides with the natural frequency of the system, resonance occurs.Below the resonance point, the object oscillates in phase with theexcitation. Above the resonance point, the oscillation lags behind theexcitation by π rad. At the resonance point, the phase delay of theoscillation is π/2 rad. Resonance is a dangerous condition in amechanical system and will produce unwanted large displacements orlead to failure.

Excitation: application of harmonic force or displacement to the system, 励振

D Translate into Japanese the following paragraph about the resonanceof an undamped single‐D.O.F. system.

When the forcing frequency ω equals the natural frequency of anundamped vibration system, the amplitude of the steady stateresponse theoretically becomes infinite. This phenomenon is calledresonance. The frequency at which this occurs is called the resonancefrequency or resonant point. The phase difference between theharmonic force and the displacement of the system is exactly ‐90o atthis frequency.

E Translate the following paragraph about forced vibration intoJapanese.

The steady‐state response is the part of the response which does notdisappear as time goes on. The transient response is the part of theresponse which disappears. The general response of the dampedsingle DOF system is written by𝑥 𝐴𝑒 cos 𝑝 𝑡 𝜙 cos 𝜔𝑡 𝛽 .

This equation contains the exponential term 𝑒 which disappears,and the steady‐state response is𝑥 cos 𝜔𝑡 𝛽 .

COLUMN: 1単位取得に必要な学習時間は1週間+α

大学設置基準によると1単位の取得に必要な学習時間は45時間とされています．1週間の労働時間の目安が40時間ですので，45時間は1週間の労働時間に1日あたり1時間を加えた時間に相当します．通常，講義科目は2単位となっていますので，一つの講義を修得するためには2週間目いっぱいの学習が必要です．この原理から考えると，1セメスター（4カ月）で取得可能な単位は，16単位ということになります．講義の数では，8科目になります．皆さんの学習時間はどうなっていますか？

• 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．

COLUMN: 振動の映像を見よう

インターネットには無料で閲覧でき，振動工学の理解に役立つ動画たくさんあります．講義では，どうもしても原理・原則の説明に終始してしまいますが，皆さんに振動工学の面白さと重要さを理解してもらうためにこれらの動画を見て欲しいです．

振動の世界 http://www.kagakueizo.org/movie/industrial/333/Youtube検索キーワード [共振, Tacoma bridge, 減衰振動, Mass damper]

Answer:

𝑓 𝑡 𝑚𝑙 𝜔 sin 𝜔𝑡Q1.a) 𝑀 𝑚 𝑥 𝑐𝑥 𝑘𝑥 𝑚𝑙𝜔 sin 𝜔𝑡b)

𝑚𝑥 𝑐𝑥 𝑘 𝑥 𝑥 0Q2.a) 𝑚𝑥 𝑐 𝑥 𝑥 𝑘 𝑥 𝑥 0b)

Q3. 𝐼𝜃 𝑐𝜃 𝑘 𝜃 𝜃 0a) 𝐼𝜃 𝑐 𝜃 𝑛𝑛 𝜃 𝑘𝜃 0b)

Q4. 𝐼𝜃 𝑐𝑙 𝜃 𝑘𝑙 𝜃 𝑓𝑙

𝑥 𝐹𝜔𝑘 𝑚𝜔 1𝑝 sin 𝑝𝑡 𝐹𝑘 𝑚𝜔 sin 𝜔𝑡

Q5.

a)

b)

𝜔 12 𝑝, 32 𝑝c) d)

𝑚𝑙 𝜃 𝑐𝑙 𝜃 𝑘𝑙 𝑚𝑔𝑙 𝜃 𝑘𝑙 𝑥Q6.

𝑥 𝑣 𝐹𝜔𝑘 𝑚𝜔 1𝑝 sin 𝑝𝑡 𝐹𝑘 𝑚𝜔 sin 𝜔𝑡𝜔 11𝑝

(𝑝 )

(𝑝 )

Q7. 𝑚 1,000kg Q8.

A. 𝜔 500,   𝜔 5500

運動方程式は𝑚𝑥 𝐹 cos 𝜔𝑡 𝑘𝑥 𝑚𝑥 𝑘𝑥 𝐹 cos 𝜔𝑡解を 𝑥 𝑋cos𝜔𝑡 と仮定すると𝑚𝜔 𝑋 𝑘𝑋 cos𝜔𝑡 𝐹 cos 𝜔𝑡

𝑚𝜔 𝑘 𝑋 𝐹両辺の係数を比較 𝑋 𝐹𝑚𝜔 𝑘振幅が2 mmより小さい𝜔を求めるには𝑋 𝐹𝑚𝜔 𝑘 2 10 m𝑚 100 kg, 𝑘 3 10 N/m, 𝐹 500 N を代入𝐹𝑚𝜔 𝑘 2 10 𝐹𝑚𝜔 𝑘 2 10

が成立する

Q10 𝑚 𝑥 𝑘 𝑥 𝑘 𝑥 𝑥𝑚 𝑥 𝑘 𝑥 𝑥 0

Amplitu

de

0

δ

𝑟 𝜔𝑝10

-π/2

-πPh

ase

2

周波数が小さいときは，振幅はほとんど静たわみに等しい

r = 1 のとき，振幅は発散する

周波数が大きくなると，振幅はどんどん小さくなる

r > 1 のとき

は，位相差はπ になる

r = 1 のとき

は，位相差はπ/2 になる

r < 1 のときは，位相差ない

Q11

Q12 10 2 rad/s

Q13

rad/s rad/s

Q14(1) 𝑚𝑙 𝜃 𝑐𝑙 𝜃 𝑘𝑙 𝜃 2𝑙𝐹cos𝜔𝑡(2) 𝜃 2𝑙𝐹𝑘𝑙 𝑚𝑙 𝜔 𝑐𝑙 𝜔 cos 𝜔𝑡 𝜙2𝐹𝑙 𝑘 𝑚𝜔 𝑐𝜔 cos 𝜔𝑡 𝜙 𝜙 Tan 𝑐𝜔𝑘 𝑚𝜔

Q14(3)

𝑘𝑚 1 2𝜁 0共振する条件は，共振周波数が 0 より大きいことであるから，

𝜁 12が，共振が生じるための減衰率の最小値となる．したがって，𝜁 𝑐2 𝑚𝑘 12 𝑐 2𝑚𝑘

Q151) 𝑚𝑥 𝑐𝑥 𝑘𝑥 𝑐𝑎𝜔sin 𝜔𝑡2) 𝑋 𝑐𝑎𝜔𝑘 𝑚𝜔 𝑐𝜔3) 𝜔 𝑘𝑚4) 𝑋 𝑎 with no respect to c value.

Q16

1) 13 𝑚𝑙 𝜃 𝑙 𝑐𝜃 𝑙2 𝑘𝜃 𝐹 𝑙cos 𝜔𝑡2) 0.029 rad

𝑥 𝑋 cos 𝜔𝑡 𝑋 sin 𝜔𝑡 として

1)に代入し，𝑋 𝑋 𝑋 とせよ．

2)をω2で微分し，その微分値が0となるようなωを求めよ．

Q171) 𝑚𝑥 𝑘𝑥 𝑎𝑘cos 𝜔𝑡2) 𝑋 𝑎1 𝜔𝑝 0.0062 6.2 mm

Q181) 𝑝 2 rad/s

2)

𝑀 11 𝜔𝑝 2𝜁𝜔𝑝

11 𝑟 2𝜁𝑟𝜔 𝑝 1 2𝜁 𝜔 | . 1.4 rad/s 𝜔 | . 1.28 rad/s

𝑀 12𝜁 1 𝜁 𝑀 . 5.0 𝑀 , . 1.74

Q183) 𝑀 𝑋𝛿 32210 12𝜁 1 𝜁 𝜁 0.4

𝜁 0.4以上となるためには c > 5.6 Ns/m

Q19 𝑋 𝐹|𝑘 𝑚𝜔 |を使う．

𝑚 2.533 … kg𝑘 42,000 N/m𝑝 128.77.. rad/s20.494. . Hz

Q201) 𝑘~2.0 10 N/m

無減衰系での力伝達率 𝑇 11 𝜔𝜔 𝑇 0.1伝達率が10%未満になるということは，ωが固有振動数よりも大きいときである

ので，1 0 である（位相が反転する）ことに留意して，11 𝜔𝜔 0.1 11 𝜔𝜔 11 2𝜋 · 17𝑘/20𝑘 2.0744 10 N/m

2) ζ > 0.177, c > 227 N・s/m

減衰系での力伝達率 𝑇 1 4𝜁 𝜔𝜔1 𝜔𝜔 4𝜁 𝜔𝜔 3

今，𝜔 𝜔 なので，1 4𝜁4𝜁 3 132 𝜁 0.17678 𝜁

Q211) 2)22.1 𝜇m 0.092 3) 33.12 N

1) 𝑋 𝐹/𝑘1 𝜔𝑝 2𝜁 𝜔𝑝

360/7600001 4.73 2 · 0.18 · 4.73

𝑝 𝑘𝑚 760000120 79.58 … 𝜔 2𝜋 360060 376.99 …𝜔𝑝 376.9979.58 4.73 …

360/7600001 4.73 2 · 0.18 · 4.73 2.209 10

Q212) 𝑇 𝐹𝐹 𝑘 𝑐 𝜔 𝑋𝐹 𝑘 𝑐 𝜔 𝑋𝑘𝛿1 2𝜁 1 2 · 0.18 · 4.73 0.092 …

𝑋 2.209 10𝛿 𝐹𝑘 4.74 … 103) 360 N 0.092 33.12 N

Q22 𝑘 𝑚𝜔3Q23 𝑣 𝜆2𝜋 𝑘𝑚

Q24(1)

𝑚𝑥 𝑐𝑥 𝑘𝑥 𝑐𝜔𝛿 cos 𝜔𝑡 𝑘𝛿 sin 𝜔𝑡𝑚𝑥 𝑐𝑥 𝑘𝑥 𝑐𝜔𝛿 sin 𝜔𝑡 𝑘𝛿 cos 𝜔𝑡𝑚𝑥 𝑐𝑥 𝑘𝑥 𝛿 𝑘 𝑐𝜔 cos 𝜔𝑡 𝜙𝜔 2𝜋𝑣𝜆

𝜙 Tan 𝑐𝜔𝛿𝑘𝛿上記の3形態のいずれも正解である．(3)の回答にもっとも容易にたどり着くのは，3番目の形態である．

(2) 𝜔 𝑘𝑚(3) 𝛿 𝑘 𝑐𝜔𝑘 𝑚𝜔 𝑐𝜔

Q24(1)

𝑚𝑥 𝑐𝑥 𝑘𝑥 𝑐𝜔𝛿 sin 𝜔𝑡 𝑘𝛿 cos 𝜔𝑡𝑚𝑥 𝑐𝑥 𝑘𝑥 𝛿 𝑘 𝑐𝜔 cos 𝜔𝑡 𝜙

T 2𝜋𝜔 𝜆𝑣

∵ 𝜙 Tan 𝑐𝜔𝑘

角周波数𝜔 ⇒ 1周期にかかる時間Tについて等式を立てる𝜔 2𝜋𝑣𝜆

詳解

基礎の調和変位を、 y 𝛿cos𝜔𝑡 とする

𝑚𝑥 𝑐 𝑥 𝑦 𝑘 𝑥 𝑦ばねの伸びは𝑥 𝑦、ダンパの両端の相対速度は𝑥 𝑦なので

𝑚𝑥 𝑐 𝑥 𝜔𝛿 sin 𝜔𝑡 𝑘 𝑥 𝛿 cos 𝜔𝑡(2) c 0なので運動方程式は 𝑚𝑥 𝑘𝑥 𝑘𝛿 cos 𝜔𝑡

固有角振動数𝑝は 𝑝 𝑚𝑘

Q24(3)

𝛿 𝑘 𝑐𝜔𝑘 𝑚𝜔 𝑐𝜔

解を 𝑥 𝑋 cos 𝜔𝑡 𝜙 𝑋 sin 𝜔𝑡 𝜙 と仮定する

運動方程式

𝑘 𝑚𝜔 𝑋 𝑐𝜔𝑋 cos 𝜔𝑡 𝜙 𝑐𝜔𝑋 𝑘 𝑚𝜔 𝑋 sin 𝜔𝑡 𝜙𝛿 𝑘 𝑐𝜔 cos 𝜔𝑡 𝜙両辺の係数を比較すると𝑘 𝑚𝜔 𝑋 𝑐𝜔𝑋 𝛿 𝑘 𝑐𝜔 𝑐𝜔𝑋 𝑘 𝑚𝜔 𝑋 0

𝑚𝑥 𝑐𝑥 𝑘𝑥 𝛿 𝑘 𝑐𝜔 cos 𝜔𝑡 𝜙

𝑋 , 𝑋 を求め、解 𝑥 に代入

𝑥 𝑘 𝑚𝜔 𝛿 𝑘 𝑐𝜔𝑘 𝑚𝜔 𝑐𝜔 cos 𝜔𝑡 𝜙 𝑐𝜔𝛿 𝑘 𝑐𝜔𝑘 𝑚𝜔 𝑐𝜔 sin 𝜔𝑡 𝜙よって求める振幅は

Q25 0.132 cos 𝜔 𝑡 0.132 0.075 sin 𝜔 𝑡 0.235 Q26

1) 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

Amplitu

de

0

δ

𝑟 𝜔𝑝10

-π/2

-π

Phase

2

周波数が小さいときは，振幅はほとんど静たわみに等しい

r ~ 1 のとき，振幅

は最大になる（発散はしない）

周波数が大きくなると，振幅はどんどん小さくなる

r > 1 のとき

は，位相差はπ に近付く

r = 1 のとき

は，位相差はπ/2 になる

Q27 R11) 𝑋 𝑎 𝑔𝑙𝑔𝑙 𝜔 2) 𝜔 2 𝑔𝑙

S1

(1) 𝑎 cos2𝜋 𝑣𝑡𝑙(2) 𝑦 𝑎 cos2𝜋 𝑣𝑡𝑙𝑦 𝑎 cos2𝜋 𝑣𝑡𝑙 とおいて，

𝑚𝑥 𝑐 𝑦 𝑥 𝑐 𝑦 𝑥 𝑘 𝑦 𝑥 𝑘 𝑦 𝑥もしくは𝑚𝑥 𝑐 𝑐 𝑥 𝑘 𝑘 𝑥 𝑐 𝑦 𝑐 𝑦 𝑘 𝑦 𝑘 𝑦

(3)

𝑚𝑥 2𝑐𝑥 2𝑘𝑥 2𝑐𝑦 2𝑘𝑦𝑦 𝑎cos2𝜋 𝑣𝑡𝑙 𝑎cos𝜔𝑡 とおくと，運動方程式は，

となる．右辺を変換し，𝑚𝑥 2𝑐𝑥 2𝑘𝑥 2𝑐𝑎𝜔sin𝜔𝑡 2𝑘𝑎cos𝜔𝑡2𝑎 𝑘 𝑐 𝜔 cos 𝜔𝑡 𝜙を得る．𝑀𝑥 𝐶𝑥 𝐾𝑥 𝐹cos𝜔𝑡 が与えられている時，xの振幅は𝐹𝐾 𝑀𝜔 𝑐𝜔 であり，共振点は

𝐾𝑀 1 2𝜁である．したがって 𝜔 2𝜋 𝑣𝑙 2𝑘𝑚 1 𝑐𝑚𝑘 から

𝑣 1𝜋 𝑘2𝑚 𝑚𝑘 𝑐𝑚𝑘 𝑙 S2(1) 𝐼 2𝑚𝑙(2) 𝐼𝜃 𝑐𝑙 𝑙𝜃 𝑥 𝑐𝑙 𝜃 𝑙 𝑘𝜃 0𝐼𝜃 2𝑐𝑙 𝜃 𝑙 𝑘𝜃 𝑙𝑐𝑋𝜔 cos 𝜔𝑡(3) c 2𝑐𝑙 𝑘′ 𝑙 𝑘 𝐹 𝑙𝑐𝑋𝜔 として𝜃 𝐹𝑘 𝐼𝜔 𝑐′𝜔 cos 𝜔𝑡 𝜙(4)

𝜙 tan 𝑐 𝜔𝑘 𝐼𝜔𝜙 𝜋/2 のときであり，𝑘 𝐼𝜔 が条件

B  減衰の無い系で振動が始まると，さらなるエネルギー入力がない限

りは，その運動は系の固有振動で維持される．加振の周波数が固有振動と同じとき，系のエネルギーの合計は外力が成す仕事によって増え，振幅の連続的な増加につながる．加振周波数が固有周波数と異なるとき，加振周波数での運動を維持するためには仕事の入力が必要である．

A 機械振動は建物の構造を通じて長距離を伝播し得る．これに関する一つ

の問題は，振動に弱い機械を建物の中で通常生じる外乱から絶縁することである．振動に弱い機械の例は，手術用顕微鏡，電子機器，レーザー，MRI装置，コンピュータの記憶ディスクなどである．伝達率を最小化する

（絶縁効果を最大化する）ためには，励振周波数が絶縁系の固有周波数よりもできるだけ大きい必要がある．減衰は共振点付近で絶大な効果を発揮し，振幅を減じる．減衰の興味深い効果は，励振周波数が 2𝑝より大きい場合は振幅が増えるという点である．

C 系に減衰が無く，励振の周波数が系の固有周波数と一致するとき，共振が生じる．共振点以下では，物体は励振と同相で振動する．共振点以上では，物体の振動は励振に π rad 遅れる．共振点では，振動は位相遅れは π/2 rad である．共振は機械系にとって危険な状態であり，好ましくない大きな変位を生じたり，破壊につながる．

D 強制振動の周波数ωが，減衰の無い振動系の固有周波数に等しい時，定常状態での応答は理論上，無限になる．この現象は共振と呼ばれる．これが生じる周波数を共振周波数または共振点という．この周波数では，調和外力と振動振幅の位相差は調度‐90oとなる．

E  定常応答とは，応答の中でも時間が経っても消失しない部分のこと

である．過渡応答とは，消失する部分のことである．一般的な減衰のある一自由度の系の応答は （式1） と書ける．この式は，消失する自然対数の底の項 𝑒 を含み，定常応答は （式2） となる．