Contoh Contoh Soal Getaran Bebas Tanpa Redaman. #1.
Satu beban 3 kg tergantung pada pegas sehingga pegas berdefleksi memanjang 42 mm. Hitung konstanta kekakuan pegas k itu. itu.
#2.
Hitung frekuensi pribadi f n dari sistim massapegas seperti pada gambar disamping.
#3.
Satu sistim massa-pegas seperti gambar disamping ditarik kebawah sejauh 100 mm dari titik posisi setimbangnya (equilibrium) kemudian dilepas. Hitung periode getarannya , defleksi statiknya pada saat sistim diam , dan kecepatan maksimumnya v max max.
st
#4.
Suatu plunyer vertikal dengan massa 2,5 kg terpasang seperti pada gambar yaitu ditahan oleh 2 buah pegas dengan konstanta kekakuan masingmasing yang dalam keadaan tertekan. Hitung frekuensi pribadi f n getaran dari plunyer ini.
#5.
Dalam keadaan setimbang silinder dengan massa 30 kg menekan pegas spiral pada gambar disamping sehingga berdefleksi tertekan sejauh 50 mm. Kemudian silinder tersebut ditekan lagi kebawah sejauh 25 mm dan dilepaskan. Hitung berapa Hz frekuensi getaran pribadi f n sistim ini kemudian bergetar.
#6.
Satu silinder dengan massa 10 kg diikat dengan pegas dan melalui pulli pulli seperti pada gambar disamping. Hitung frekuensi pribadi f n sistim ini bila bergetar.
#7.
Suatu peluncur dengan massa 0,1 kg berosilasi didalam lubang. Peluncur itu disebelah kanan ditahan oleh 1 buah pegas, dan di sebelah kiri ditahan oleh 2 buah pegas yang masing mempunyai konstanta kekakuan pegas
k = 90 N/m. Pada saat
awal t = 0 simpangan x0 = 3 mm d an kecepatannya v = 12 mm/detik. Hitung besarnya simpangan dan kecepatannya setelah waktu t = 2 detik. Hitung pula periode getarannya .
#8.
Tentukan konstanta kekakuan pegas eqivalen k eq pengganti pegas pegas pada gambar (a) dan (b) disamping ini.
#9.
Seorang wanita dengan berat massa 55 kg berdiri ditengah papan sehingga papan tersebut berdefleksi ditengah sejauh 22mm. Bila wanita itu bergerak sehingga papan bergetar, hitung berapa frekuensi pribadi f n getaran papan tersebut.
#10.
Massa sebesar 25 kg tergantung pada 2 buah pegas melalui katrol atau pulli dengan masing masing konstanta kekakuan k = 1200 N/m. hitung frekuensi pribadi f n getaran dari sistim ini.
#11.
Hitung frekuensi pribadi f n getaran dari sistim katrol seperti pada gambar disamping.
#12.
Hitung frekuensi pribadi f n getaran dari sistim katrol seperti pada gambar disamping.
A rt ic le 8 / 2
F re e V ib ra ti on o f P ar ti cl es
609
Sample Problem 8/1 A body weighing 25 lb is suspended from a spring of constant k 160 lb/ft. At time t 0, it has a downward velocity of 2 ft/sec as it passes through the position of static equilibrium. Determine
k = 160 lb /ft
(a) the static spring deflection st (b) the natural frequency of the system in both rad/sec (n) and cycles/sec ( ƒn )
W = 25 lb
( c) the system period (d) the displacement x as a function of time, where x is measured from the position of static equilibrium
F s = kδ st
( e) the maximum velocity vmax attained by the mass
k(δ st + x)
δ st
( ƒ) the maximum acceleration amax attained by the mass.
Equilibrium position
kx
x ≡
Solution.
(a) From the spring relationship F s kx, we see that at equilibrium
mg kst
(b)
mg 25 0.1562 ft or 1.875 in. 160 k
st
n
k m
n
( c)
1 ƒ
n
mg
Ans. Helpful Hints
160 14.36 rad/sec 25/32.2
1 ƒ (14.36) 2
mg
2.28
Ans.
You should always exercise extreme caution in the matter of units. In the subject of vibrations, it is quite easy to commit errors due to mixing of feet and inches, cycles and radians, and other pairs which frequently enter the calculations.
Ans.
cycles/sec
1 0.438 sec 2.28
Ans.
(d) From Eq. 8/6:
Recall that when we refer the mo-
x x0 cos t
x ˙0
n
(0)
n
sin
cos 14.36 t
0.1393
tion to the position of static equilibrium, the equation of motion, and therefore its solution, for the present system is identical to that for the horizontally vibrating system.
nt
2 sin 14.36 t 14.36
sin 14.36 t
Ans.
As an exercise, let us determine x from the alternative Eq. 8/7:
x x02 ( x ˙0 / )2 sin n
02
0.1393
2 14.36
nt tan1
2
˙ x0
n
x 0
sin 14.36t tan1
(0)(14.36) 2
sin 14.36 t
( e) The velocity is x ˙ 14.36(0.1393) cos 14.36 t 2 cos 14.36t. Because the cosine function cannot be greater than 1 or less than 1, the maximum velocity vmax is 2 ft/sec, which, in this case, is the initial velocity. Ans. ( ƒ) The acceleration is
x ¨ 14.36(2) sin 14.36 t 28.7 sin 14.36 t The maximum acceleration amax is 28.7 ft/sec 2.
Ans.
Sample Problem 8/3 The two fixed counterrotating pulleys are driven at the same angular speed 0. A round bar is placed off center on the pulleys as shown. Determine the natural frequency of the resulting bar motion. The coefficient of kinetic friction between the bar and pulleys is k.
ω 0
ω 0
a Central
y
Sample Problem 8/3 The two fixed counterrotating pulleys are driven at the same angular speed 0. A round bar is placed off center on the pulleys as shown. Determine the natural frequency of the resulting bar motion. The coefficient of kinetic friction between the bar and pulleys is k.
ω 0
a –– 2
The free-body diagram of the bar is constructed for an arbitrary displacement x from the central position as shown. The governing equations are k N A k N B mx ¨
[Σ F y 0]
N A N B mg 0
[Σ M A 0]
µ k N A
G
A
B
µ k N B
N B
mg
Because the bar is slender and does not rotate, the use of a moment equilibrium equation is justified.
2 k g x ¨ a x 0 We recognize the form of this equation as that of Eq. 8/2, so that the n atural frequency in radians per second is n 2 k g / a and the natural frequency in cycles per second is ª n
a –– 2
Helpful Hints
Eliminating N A and N B from the first equation yields
x
N A
a x mg 0 2
aN B
y
Central position
Solution.
[Σ F x mx ¨]
ω 0
a
1 2 k g / a 2
We note that the angular speed
0 does not enter the equation of motion. The reason for this is our assumption that the kinetic friction force does not depend on the relative velocity at the contacting surface.
Ans.
Neutral position
Sample Problem 8/5
x B = b cos
The spring attachment point B is given a horizontal motion x B b cos t. Determine the critical driving frequency c for which the oscillations of the mass m tend to become excessively large. Neglect the friction and mass associated with the pulleys. The two springs have the same stiffness k.
ω t
k
B T T
Equilibrium position
Solution. The free-body diagram is drawn for arbitrary positive displacements x and x B. The motion variable x is measured downward from the position of static equilibrium defined as that which exists when x B 0. The additional stretch in the upper spring, beyond that which exists at static equilibrium, is 2 x x B. Therefore, the dynamic spring force in the upper spring, and hence the dynamic tension T in the cable, is k(2 x x B). Summing forces in the x-direction gives
x
m
k
kx
[Σ F x mx ¨]
(Dynamic forces only)
2 k(2 x x B) kx mx ¨
which becomes
Helpful Hints x ¨
The natural frequency of the system is c
636
C ha pt er 8
If a review of the kinematics of con-
5 k 2 kb cos t x m m n
n
strained motion is necessary, see Art. 2/9.
5 k / m. Thus,
5 k / m
V ib ra ti on a nd T i me R es po ns e
We learned from the discussion in Ans.
Art. 8/2 that the equal and opposite forces associated with the position of static equilibrium may be omitted from the analysis. Our use of the terms dynamic spring force and dynamic tension stresses that only the force increments in addition to the static values need be considered.
Neutral position x B = b cos ω t
Sample Problem 8/5
k
The spring attachment point B is given a horizontal motion x B b cos t. Determine the critical driving frequency c for which the oscillations of the mass m tend to become excessively large. Neglect the friction and mass associated with the pulleys. The two springs have the same stiffness k.
B T T
Equilibrium position Solution.
The free-body diagram is drawn for arbitrary positive displacements x and x B. The motion variable x is measured downward from the position of static equilibrium defined as that which exists when x B 0. The additional stretch in the upper spring, beyond that which exists at static equilibrium, is 2 x x B. Therefore, the dynamic spring force in the upper spring, and hence the dynamic tension T in the cable, is k(2 x x B). Summing forces in the x-direction gives
m
x
k
kx
[Σ F x mx ¨]
(Dynamic forces only)
2 k(2 x x B) kx mx ¨
which becomes
Helpful Hints
The natural frequency of the system is c
636
C ha pt er 8
If a review of the kinematics of con-
5 k 2 kb cos t x m m
x ¨
n
strained motion is necessary, see Art. 2/9.
5 k / m. Thus,
We learned from the discussion in
5 k / m n
Art. 8/2 that the equal and opposite forces associated with the position of static equilibrium may be omitted from the analysis. Our use of the terms dynamic spring force and dynamic tension stresses that only the force increments in addition to the static values need be considered.
Ans.
V ib ra ti on a nd T i me R es po ns e
Sample Problem 8/7
O y
A simplified version of a pendulum used in impact tests is shown in the figure. Derive the equation of motion and determine the period for small oscillations about the pivot. The mass center G is located a distance r 0.9 m from O, and the radius of gyration about O is kO 0.95 m. The friction of the bearing is negligible.
O x
O –
–
r
r
Solution. We draw the free-body diagram for an arbitrary, positive value of the
angular-displacement variable , which is measured counterclockwise for the coordinate system chosen. Next we apply the governing equation of motion to obtain
[Σ M O I O ¨]
θ
θ
G
mgr sin mkO2 ¨
¨
or
gr sin 0 kO2
Ans. mg
Note that the governing equation is independent of the mass. When sin , and our equation of motion may be written as
¨
gr kO
2
is small, Helpful Hints
With our choice of point O as the
0
moment center, the bearing reactions O x and O y never enter the equation of motion.
The frequency in cycles per second and the period in seconds are ƒn
For the given properties:
1 2
gr
kO2
1 ƒn
2
2
kO2 gr
(0.95)2 2.01 s (9.81)(0.9)
Ans.
Ans.
For large angles of oscillation, determining the period for the pendulum requires the evaluation of an elliptic integral.
l
— 2
l
— 2
636
C ha pt er 8
V ib ra ti on a nd T i me R es po ns e
Sample Problem 8/7
O y
A simplified version of a pendulum used in impact tests is shown in the figure. Derive the equation of motion and determine the period for small oscillations about the pivot. The mass center G is located a distance r 0.9 m from O, and the radius of gyration about O is kO 0.95 m. The friction of the bearing is negligible.
O x
O –
–
r
r
Solution. We draw the free-body diagram for an arbitrary, positive value of the
angular-displacement variable , which is measured counterclockwise for the coordinate system chosen. Next we apply the governing equation of motion to obtain
[Σ M O I O ¨]
θ
θ
G
mgr sin mkO2 ¨
¨
or
gr sin 0 kO2
Ans. mg
Note that the governing equation is independent of the mass. When sin , and our equation of motion may be written as
¨
gr kO
2
is small, Helpful Hints
With our choice of point O as the
0
moment center, the bearing reactions O x and O y never enter the equation of motion.
The frequency in cycles per second and the period in seconds are ƒn
1 2
gr
kO2
For the given properties:
1 ƒn
2
2
kO2 gr
(0.95)2 2.01 s (9.81)(0.9)
Ans.
Ans.
For large angles of oscillation, determining the period for the pendulum requires the evaluation of an elliptic integral.
l
l
— 2
Sample Problem 8/8
— 2
O
m
The uniform bar of mass m and length l is pivoted at its center. The spring of constant k at the left end is attached to a stationary surface, but the right-end spring, also of constant k, is attached to a support which undergoes a harmonic motion given by y B b sin t. Determine the driving frequency c which causes resonance.
k
k y B = b sin
ω t
B
Helpful Hints
As previously, we consider only the Solution. We use the moment equation of motion about the fixed point O to
obtain
1 l l l l k sin cos k sin y B cos ml2 ¨ 2 2 2 2 12
Assuming small deflections and simplifying give us
¨
6 k m
6 kb sin t ml
The natural frequency should be recognized from the now-familiar form of the equation to be
6 k / m
Thus, c n 6 k / m will result in resonance (as well as violation of our smallangle assumption!). Ans.
C ha p te r 8
O y θ
k
O x
(
l
V ib ra ti on a nd T im e Re s po ns e
Sample Problem 8/10
)
— sin θ 2
mg
k
( l
)
— sin θ – y B 2
The standard form here is ¨ n2 M 0 sin t
n
646
changes in the forces due to a movement away from the equilibrium position.
klb and I O 2 I O 1 ml2. The natural frequency n of a 12 system does not depend on the external disturbance. , where M 0
l
l
— 2
Sample Problem 8/8
— 2
O
m
The uniform bar of mass m and length l is pivoted at its center. The spring of constant k at the left end is attached to a stationary surface, but the right-end spring, also of constant k, is attached to a support which undergoes a harmonic motion given by y B b sin t. Determine the driving frequency c which causes resonance.
k
k y B = b sin
B
ω t
Helpful Hints
As previously, we consider only the Solution. We use the moment equation of motion about the fixed point O to
obtain
1 l l l l k sin cos k sin y B cos ml2 ¨ 2 2 2 2 12
O y
Assuming small deflections and simplifying give us
¨
6 k m
θ
6 kb sin t ml
O x
k
The natural frequency should be recognized from the now-familiar form of the equation to be
(
l
)
— sin θ 2
k
mg
( l
)
— sin θ – y B 2
The standard form here is ¨ n2 M 0 sin t
n
6 k / m
Thus, c n 6 k / m will result in resonance (as well as violation of our smallangle assumption!). Ans.
646
changes in the forces due to a movement away from the equilibrium position.
C ha p te r 8
klb , where M 0 and I O 2 I O 1 ml2. The natural frequency n of a 12 system does not depend on the external disturbance.
V ib ra ti on a nd T im e Re s po ns e
Sample Problem 8/10 The small sphere of mass m is mounted on the light rod pivoted at O and supported at end A by the vertical spring of stiffness k. End A is displaced a small distance y0 below the horizontal equilibrium position and released. By the energy method, derive the differential equation of motion for small oscillations of the rod and determine the expression for its natural frequency n of vibration. Damping is negligible.
b O
O
2
Here again, we note the simplicity of
where we see that the vertical displacement of m is (b/l) y. Thus, with the energy sum constant, its time derivative is zero, and we have
d d 1 b (T V ) m y ˙ l dt dt 2
2
1 2 ky 2
0
which yields y ¨
Helpful Hints motion of the end of the bar would cause our expression for the deflection of the spring to be in error.
l2 k y 0 b2 m
δ st
A
For large values of y, the circular
The kinetic energy in the displaced position is b 1 m y ˙ 2 l
k
Equilibrium position
1 V ky2 2
T
b — y l m
where st is the static deflection of the spring at equilibrium. But the force in the spring in the equilibrium position, from a zero moment sum about O, is (b / l)mg kst. Substituting this value in the expression for V and simplifying yield
A
y
b 1 1 V V e V g k( y st)2 kst2 mg y 2 2 l
m l
Solution. With the displacement y of the end of the bar measured from the
equilibrium position, the potential energy in the displaced position for small values of y becomes
k
Ans.
the expression for potential energy when the displacement is measured from the equilibrium position.
646
C ha p te r 8
V ib ra ti on a nd T im e Re s po ns e
Sample Problem 8/10 The small sphere of mass m is mounted on the light rod pivoted at O and supported at end A by the vertical spring of stiffness k. End A is displaced a small distance y0 below the horizontal equilibrium position and released. By the energy method, derive the differential equation of motion for small oscillations of the rod and determine the expression for its natural frequency n of vibration. Damping is negligible.
b O
O
2
Here again, we note the simplicity of
where we see that the vertical displacement of m is (b/l) y. Thus, with the energy sum constant, its time derivative is zero, and we have
d d 1 b (T V ) m y ˙ l dt dt 2
2
1 2 ky 2
0
which yields l2 k y 0 b2 m
Ans.
when y ˙ is canceled. By analogy with Eq. 8/2, we may write the motion frequency directly as n
l k / m b
Ans.
Alternatively, we can obtain the frequency by equating the maximum kinetic energy, which occurs at y 0, to the maximum potential energy, which occurs at y y0 ymax, where the deflection is a maximum. Thus, gives
T max V max
b 1 m y ˙ 2 l max
2
1 ky 2 2 max
Knowing that we have a harmonic oscillation, which can be expressed as y ymax sin nt, we have y ˙max ymaxn. Substituting this relation into our energy balance gives us
b 1 m ymaxn 2 l
2
1 ky 2 2 max
Helpful Hints motion of the end of the bar would cause our expression for the deflection of the spring to be in error.
b 1 m y ˙ 2 l
so that
as before.
JURUSAN TEKNIK MESIN FAKULTAS TEKNIK U.I.
n
l k / m b
δ st
A
For large values of y, the circular
The kinetic energy in the displaced position is
y ¨
k
Equilibrium position
1 V ky2 2
T
b — y l m
where st is the static deflection of the spring at equilibrium. But the force in the spring in the equilibrium position, from a zero moment sum about O, is (b / l)mg kst. Substituting this value in the expression for V and simplifying yield
A
y
b 1 1 V V e V g k( y st)2 kst2 mg y 2 2 l
m l
Solution. With the displacement y of the end of the bar measured from the
equilibrium position, the potential energy in the displaced position for small values of y becomes
k
Ans.
the expression for potential energy when the displacement is measured from the equilibrium position.
JURUSAN TEKNIK MESIN FAKULTAS TEKNIK U.I.
Lengan AB dari keran pengangkat muatan di kapal terbuat dari batang baja uniform dengan panjang 10 m dan luas penampang 2500 mm². Satu muatan yang akan dimuat kedalam kapal tergantung statis pada keran dengan berat W. Kabel Penarik beban dan penahan CDEBF terbuat dari baja pula dengan luas penampang 100 mm². Rangkaian kabel CDEB pada pulley bisa diabaikan, dianggap satu kabel saja. Hitung konstanta elastisitas (pegas ekivalen) dari sistim pengangkat muatan ini pada arah vertikal. Modulus Young baja E = 207·109 N/m².
Satu ocean-going luxury liner yang besar mempunyai masalah getaran. Kapal ini mempunyai 4 propeller dengan massa 12.200 kg setiap propellernya. Masing masing propeller digerakkan oleh poros berrongga yang panjang dengam ukuran L= 71,6 m, ID= 0,28 m dan OD= 0,56 m. Kapal berlayar dengan putaran poros adalah 258 rpm. Hitung frekuensi pribadi getaran longitudinal dari poros panjang dan propeller tersebut. Apa yang terjadi bila propeller yang digunakan mempunyai 4 sudu. Bagaimana cara mengkoreksi masalah getaran tersebut.
Satu pesawat terbang ringan dengan propulsi baling baling digerakkan oleh motor bakar 4-langkah 9 silinder radial. Daya motor diteruskan melalui satu poros pendek kepada kedua bilah baling baling yang terbuat dari aluminium. Momen inersia baling baling arah aksial besarnya adalah 17,62 kg m2, sedangkan momen inersia efektif dari seluruh komponen mesin radial yang bergerak/berputar adalah 0,544 kg m2. Konstanta elastisitas puntir dari poros baling baling adalah 0,45 ×106 N m2/rad. (a) Hitung berapa besarnya frekuensi pribadi dari sistim propulsi pesawat terbang tersebut. (b) Bila motor radial penggerak itu dioperasikan pada kecepatan putaran 200 rpm, jelaskan apa yang kemungkinan terjadi. •
•
•
Satu model rudder pesawat terbang bila dipandang dari atas adalah seperti pada gambar disamping, dan dapat dianggap sebagai sebagai satu batang kaku dengan engsel pegas torsi Kt . Bila tahanan udara pada kedua sisi rudder dapat dianggap sebagai pegas linier k 1 dan k 2, sedangkan Xsinω t aliran udara turbulensi pada rudder dianggap menimbulkan eksitasi getaran paksa dengan respons simpangan Xsinω t dan defleksi rudder adalah sudut yang kecil maka hitunglah frekuensi pribadi d ari rudder ini.
King Fahd University of Petroleum & Minerals Mechanical Engineering Department (Semester 061)
ME 482 Mechanical Vibrations
HW # 1 - Solution
( Assigned: Monday, 5 March 2007 Due: Monday, 12 March 2007) _____________________________________________________________________________ 1) Write the equations of motion (EOM) and then find the natural frequency for the following inclined mass-spring system.
2) Find the EOM, natural frequency and damping ratio for the following system.
