MCQ MDL 1A
FREE UNDAMPED VIBRATIONS
Q.1 An automobile having a mass 1500 kg deflects its spring by 5 cms under its dead weight. The
natural frequency of the car in vertical direction in Hz will be,
a) 2.229
b) 4.29 6
c) 3.784
d) 5.626
Q.2 Spring mass system having spring stiffness 'K' N/m and mass of 'M' kg has its natural
frequency of vibration 12 Hz. on adding 2 kg. mass ,frequency reduces by 2 Hz. Then the mass in
kg will be,
a) 5.64
b) 4.54
c) 6.62
d) 7.34
Q.3 Spring mass system having spring stiffness 'K' N/m and mass of 'M' kg has its natural
frequency of vibration 12 Hz. on adding 2 kg. mass ,frequency reduces by 2 Hz. Then the stiffness
in N/m will be,
a) 25808.9
b) 29876.9
c) 21456.7
d) 22123.8
Q.4 A body performs two vibratory motion simultaneously. X1 = 1.93 sin 9.5 t
X2 = 2.00 sin 10.0 t ,units being cm., rad. and sec. The maximum beat amplitude and the beat in
m and rad/sec frequency will be
a) 0.7 , 19.5 b) 3.93, 19.5 c) 3.93 , 0.5 d) 0.07 , 0.5
Q.5
a) 2
Degree of freedom of a ship sailing will be
b) 3
c) 4
d) 6
Q.6 A body suspended from a spring vibrates vertically up and down between two positions
3cms and 5 cms above ground. During each sec. it reaches the top position(5cm above ground) 15
times. Its time period in sec. and amplitude in cm. of above body will be ,
a) 0.06 , 1
b) 0.08 , 1.5 c) 0.07 , 2
d) 0.07 , 3
Q.7 In all wall clocks which work on the principle of simple pendulum, the length of string in
mts. that can be worked out for a time period of 1 sec. will be,
a) 0.6495
b) 0.2485
c) 0.4976
d) 0.5219
Q.8 If two vibrations of same frequency are such that when one reaches its maximum
displacement (peak acceleration point) the displacement of the other is zero. Phase difference
between the two vibration is,
a) π/6
b) π/4
c) π/2
d) π/8
Q.9 The direction of inertia force of piston of a vertical engine when piston
moves from mid-position to T.D.C. Will be
a) Downwards
b)Left
c) Right
d) Upwards
Q.10 Four balls of mass 2 kg. , 3 kg. , 5kg. and 6 kg. respectively are suspended using strings of
equal lengths. The time period of oscillations will be ,
a) Same in the all the cases
b) Max. in the case of 2 kg. mass
c) Max. in the case of 6 kg. mass
d) Min. in the case of 2 kg. mass
Q.11 A 5 kg. body is suspended from a spring of constant 'K' . K = 1000 N/mm .Mass of the
spring is 0.6 kg. Natural frequency of the above system considering the mass of the system in kg
will be,
a) 3.032
b) 2.207
c) 4.702
d) 8.302
MCQ MDL 1B
Q.1 A car having a mass 1270 kg deflects its spring by 4 cms under its dead weight. The natural
frequency of the car in Hz in vertical direction will be,
2.492
4.296 3.784
5.626
Q.2 Spring mass system having spring stiffness 'K' N/m and mass of 'M' kg has its natural
frequency of vibration 10 Hz. on adding 2 kg. mass ,frequency reduces by 2 Hz. Then the mass in
kg will be,
5.64
3.555
6.62
7.34
Q.3 Spring mass system having spring stiffness 'K' N/m and mass of 'M' kg has its natural
frequency of vibration 10 Hz. on adding 2 kg. mass ,frequency reduces by 2 Hz. Then the stiffness
in N/m will be,
14034.57
29876.9
21456.7
22123.8
Q.4 A body performs two vibratory motion simultaneously. X1 = 1.90 sin 9.1 t
X2 = 2.80 sin 11.8 t ,units being cm., rad. and sec. The maximum beat amplitude in meters and
the beat frequency in rad/sec will be
2.71 , 19.5
3.93, 19.5
4.70 , 2.70
3.07 , 0.5
Q.5 A torsional vibrating having shaft stiffness 'Kt' N-m/rad and mass moment of inertia of
'I' Kg-m2 has its natural frequency of vibration 15 Hz. On doubling the mass moment of inertia
of 'I'2 and reducing the shaft stiffness 'Kt' to it’s half, the new frequency in Hz
will be,
5.64
7.50
6.62
7.34
Q.6 A body suspended from a spring vibrates vertically up and down between two positions
2cms and 6 cms above ground. During each sec. it reaches the top position(6cm above ground) 12
times. Its time period in sec. and amplitude in cm. of above body will be ,
0.08 , 2 0.08 , 1.5 0.07 , 2
0.07 , 3
Q.7 A torsional vibrating having shaft stiffness 'Kt' N-m/rad and mass moment of inertia of
'I' Kg-m2 has its natural frequency of vibration 12.6 Hz. On doubling the shaft stiffness 'Kt' and
reducing the mass moment of inertia of 'I'2 to it’s half, the new frequency in Hz
will be,
45.6
25.2 28.4
35.2
Q.8 If two tides of same frequency would reach the sea bank with a time lag of quarter time
period (tp) Phase difference between the two will be,
π/6
π/4
π/2
π/8
Q.9 The direction of inertia force of piston of a vertical engine when piston
moves from mid-position to B.D.C. Will be
Downwards
Left
Right Upwards
Q.10 A machine is installed on four springs each of stiffness K = 33 N/m .Mass of the machine is
2.8 kg. Natural frequency of the above system in Hz will be,
2.032 1.09
2.702 3.302
DAMPED FREE VIBRATION
Q.1 In a under damped system with a mass of 400 Kg. Amplitude was found to
be reducing by values measured on one side of equilibrium are
4.87,1.62,.54,.18 ,.06 etc .respectively. Number of cycles required for the
amplitude to reduce to zero is
Zero
Infinity
Four
Five
Q.2 In large guns the type of damping used will be
Over-damping
Under-damping Critical damping
options
None of the
Q.3 A horizontal spring mass system with coulomb damping has mass of 5 Kg.
attached to spring of stiffness 980 N/m and frictional resistance is 1.225 N. The
amplitude lost after 7 cycles will be
0.56 m
0.47 m
0.43 m 0.35 m
Q.4 Most of the dampers used in engineering applications use
No damping
Over damping Critical damping
Under damping
Q.5 A system has the following data m = 10 Kg., K = 2400 N/m and ζ =
0.249.Mas of the spring is 3 kg Damped frequency of the system considering
mass of the spring will be
3.29 Hz.
2.93 Hz
2.26 Hz
3.92 Hz.
Q.6 Which type of vibrations are also known as transient vibrations?
Undamped vibrations Damped vibrations Torsional vibrations Transverse
Vibrations
Q.7 The logarithmic decrement if damping factor is 0.33 will be
4.39 2.19 5.29
None of the options
Q.8 In a damped vibration system the mass having 20 kg makes 40 oscillations in
20 seconds. The amplitude of vibrations decreases to 1/8 th of its initial value after
8 oscillations. The damped frequency of vibration will be
1.2 Hz
1.4 Hz
1.6 Hz
None
Q.9 The natural frequency of damped vibration, if damping factor is 0.52 and
natural frequency of the system is 30 rad/sec which consists of machine
supported on springs and dashpots will be
35.78 rad/sec 25.62 rad/sec 38.82 rad/sec None of the options
Q.10 The logarithmic decrement, if the amplitude of a vibrating body reduces to
1/6th in two cycles will be
0.4763 0.8958 0.7839
None of the options
Q.11 The logarithmic decrement, if the amplitude of a vibrating body reduces to 1/8 th
of it’s intial value in eight cycles will be
0.2599 0.8958 0.7839
None of the options
Q.12 A gun barrel of mass 500kg has a recoil spring of stiffness 3,00,000
N/m. If the
barrel recoils 1.2 meters on firing, initial velocity of the barrel in m/sec
will be
29.39
21.43
32.37
36.73
Q.13 A gun barrel of mass 500kg has a recoil spring of stiffness 3,00,000
N/m. If the barrel recoils 1.2 meters on firing, the critical damping
coefficient of the dashpot
in N-sec/m will be
24495
19873
31687
41765
Q.14 A disc of a torsional pendulum has a moment of inertia of 600 kg-m2
and is immersed in a viscous fluid. The shaft attached to it is 0.4m long
and 0.1m in diameter. When the pendulum is oscillating, the observed
amplitudes on the same side of the mean position for successive cycles
are 9 degrees, 6 degrees and 4 degrees. The logarithmic decrement will
be
0.405 0.895 0.783
None of the options
Q.15 The logarithmic decrement if damping factor is 0.03 will be
0.188
0.895 0.783
None of the options
Q.16 A damped vibration system uses viscous damping. The amplitude
Would reduce by
Constant ratio Constant value
variable value
None of the options
Q.17 A damped vibration system has Coulomb’s damping. The amplitude
Would reduce by
Constant ratio Constant value
variable value
None of the options
Q.18 In most engineering applications the type of damping used will be
Over-damping Under-damping
the options
Critical damping
None of
Q.19 The natural frequency of damped vibration, if damping factor is 0.52 and natural
frequency of the system is 30 rad/sec will be
25.62 rad/sec 20.78 rad/sec
14.4 rad/sec
15.33 rad/sec
Q.20 In a under damped system with a mass of 600 Kg. Amplitude was
found to be reducing by values measured on one side of equilibrium are
4.87,1.62,.54,.18 ,.06 etc .respectively. Number of cycles required for the
amplitude to reduce to zero is
Zero
Infinity
Four
Five
FORCED VIBRATIONS
1. In forced vibrations the magnitude of damping force at resonance equals
Inertia force Impressed force Infinity
Spring force
In steady state forced vibrations, the amplitude of vibrations at resonance
is
Directly proportional to the damping coefficient
Directly proportional to the resonant frequency
Inversely proportional to the damping coefficient
Inversely proportional to the resonant frequency
The ratio of the maximum displacement of the forced vibration to the zero
frequency deflection is called
Logarithmic decrement Damping coefficient Critical damping coefficient
Magnification factor
2. For steady state forced vibrations, the phase lag at resonance condition in
degrees is
0
45 90 180
3. A rotary system having damping coefficient of 50N sec m/rad, when twisted
with an angular velocity of 2 rad/sec will experience the damping torque of
50 Nm
100 Nm
25 Nm
200 Nm
4. Which of the following effects is more dangerous for a ship
Rolling
Waving
Pitching
Steering
5. The critical speed of a vehicle which moves on a road having sinusoidal
profile of wavelength 2.5 m if the mass of the vehicle is 300 kg and natural
frequency of its spring suspension system is 8 rad/sec will be
4.15 m/sec 3.18 m/sec
2.36 m/sec
None of the above
6. The effect of damping on phase angle at resonance frequency will be
Phase angle increases as damping increases Damping has no effect on phase
angle Phase angle increases as damping decreases None Of The Options
7. The damped natural frequency, if a spring mass damper system is subjected
to periodic disturbing force of 30 N. Damping coefficient is equal to 0.76
times of critical damping coefficient and undamped natural frequency is 5
rad/sec will be
3.99 rad/sec
2.13 rad/sec
4.12 rad/sec
3.24 rad/sec
8. When frequency ratio (ω/ωn) is greater than unity, phase angle decreases as
______
Damping Factor Increases Damping Factor Decreases
constant None Of The Options
Damping Factor remains
9. Magnification factor is the ratio of ______
Zero frequency deflection and amplitude of steady state vibrations
Amplitude of steady state vibrations and zero frequency deflection
None of the options
10. The equation m(d2x/ dt2) + c (dx/dt) + Kx = F0 sin ωt is a second order
differential equation. The solution of this linear equation is given as
Complementary function
Particular function
Sum of complementary and particular function
Difference of complementary and particular function
11. The phase difference or phase angle in forced vibrations means
Difference between displacement vector (xp) and velocity vector Vp
Angle in which displacement vector leads force vector by (F0 sinωt)
Angle in which displacement vector (xp) lags force vector (F0 sinωt)
None Of The Options
MCQ SET 4
BALANCING ROTATING MASSES
Minimum number of balance masses required to achieve static balance of number of masses
rotating in the same plane will be
Two
Four
Six
One
The unbalanced force due to revolving masses
Varies in magnitude but constant in direction
Varies in direction but constant in magnitude
Varies in magnitude and direction both
Constant in magnitude and direction both
The unbalance in a rotor is produced due
Centripetal forces Centrifugal forces Inertia forces
Resistive forces
Minimum number of balancing planes required to achieve dynamic balance of number of masses
rotating in non-coplanar planes will be
Two
Four
Six
One
In a rotating component the effect of any unbalance located at some radius is to
Bend the shaft Compress the shaft Twist the shaft Shear the shaft
In order to achieve the complete balance of number of masses rotating in non-coplanar planes
The algebraic summation of all the forces must be zero
The algebraic summation of all the couples must be zero
The algebraic summation of all the forces and couples must be zero
None of the options
A mass m attached to a shaft rotating at ω rad/sec at radius r cm from axis of the shaft is
balanced by mass β at radius b cm from the axis of shaft . If the speed of the shaft is doubled ,
the value of balance mass β will be
Doubled
Quadrupled
Unaffected
Halved
A mass of 12 kg is attached to a shaft radius at 80cm. The balance masses B1 and B2 are
attached at a radius of 80 cm. The planes of rotation of the three masses are parallel. If the shaft
is rotating at 100 rpm and to achieve the dynamic balance, the balance mass B2 should be
9 kg
6 kg
12kg
4 kg
The balance masses are introduced in the plane parallel to the plane of rotation of the disturbing
mass. To obtain complete dynamic balance, minimum number of balance masses to be
introduced will be
One
Two
Three
None of the options
A mass of 16 kg is attached to a shaft radius 60cm. The balance masses B1 and B2 are attached
at a radius of 60 cm. The planes of rotation of the three masses are parallel. If the shaft is
rotating at 100 rpm and to achieve the dynamic balance, the balance mass B1 is
9 kg
12 kg
24 kg
18 kg
The static balancing is satisfactory for low speed rotors but with increasing speeds, dynamic
balancing becomes necessary. This is because, the
Unbalanced couples are caused only at higher speeds
Unbalanced forces are not dangerous at higher speeds
Effects of unbalances are proportional to the square of the speed
Effects of unalances are directly proportional to the speed
MCQ MDL 5
BALANCING OF REC COMPONENTS
Q.1 A single cylinder reciprocating engine has speed 240 rpm, stroke 300 mm,
mass of reciprocating parts 50 kg, mass of revolving parts at 150 mm radius
37kg. If two-third of the reciprocating parts and all the revolving parts are to be
balanced, then balancing mass required at a radius of 400 mm will be
26.38 kg
27.38 kg
28.38 kg
None of the options
Q.2 The primary unbalanced force is maximum when the angle of crank with the
line of stroke is
45 degrees
90 degrees 135 degrees
180 degrees
Q.3 The frequency of fluctuations of secondary inertia forces of reciprocating
parts of a multi cylinder engine in one revolution of crank shaft will be
Two
Four
Six
One
Q.4 Multi-cylinder engines are desirable because
Balancing problems are reduced
No balancing is required
Efficiency is enhanced
None of the options
Q.5 The secondary unbalanced force is maximum when the angle of crank with
the line of stroke is
45 degrees
90 degrees
135 degrees
None of the options
Q.6 The partial balancing means
Balancing partially the revolving masses
Balancing partially the reciprocating masses
Best balancing of engines
None of the options
Q.7 Secondary forces in reciprocating mass on engine frame are
Of same frequency as of primary forces
Twice the frequency as of primary forces
Four times the frequency of primary forces
None of the options
Q. 8 Number of times the primary unbalanced force is maximum in one
revolution of the crank
Two times Three times Four times Six times
Q.9 In case of low speed reciprocating engines ,the following is used
Complete balancing
Partial balancing Dynamic balancing Static balancing
Q.10 In multi cylinder inline engines, the balance weights in order to achieve
complete primary balance are placed on auxiliary drives rotating at
Same speed as main crank shaft Half the speed of the main crank shaft
Twice the speed of the main crank shaft
Quadraple times than main
Crank shaft
MCQ MDL 6
TORSIONAL VIB
Q.1 In a torsional vibrating system when the rotor A, the rotor B, the
rotor C, rotate in the directions shown in Fig. The number of nodes
formed will be
Zero nodes One node Two nodes Three nodes
Q.2 For a ship “ The critical speed speed range” or “The barred speed
speed range “ is evaluated based on
The axial vibrations
The transverse vibrations
The longitudinal vibrations
The torsional vibrations
Q.3 A torsional vibrating system carrying four rotors and has a degree
Of freedom(DOF) of four. It’s four natural frequencies are fn0= 0 Hz,
fn1= 27 Hz, fn2= 134 Hz, fn3= 278 Hz. When this system was vibrating,
three nodes were observed to be formed. Then in that case the might be
vibrating at
0 Hz
27 Hz
134 Hz
278 Hz
Q.4 In case of a torsional vibrating system, the internal particles of the
system would be found to be moving in
Along the lines parallel to the axis of the vibrating system
Along the lines Perpendicular to the axis of the vibrating system
In concentric circles about the axis of the vibrating system
None of the options
Q.5 The torsional vibrations are more important to be considered on
board because
They are not visible like other vibrations and felt to be evident only after
the failure of the component
They induce more stresses than other vibrations
They have higher frequencies than other vibrations
They have higher magnitudes than other vibrations
Q.6 In the torsional vibrating system when the rotor A, the rotor B, the
rotor C, rotate in the same direction shown in Fig. The number of nodes
formed will be
Zero nodes One node Two nodes Three nodes
Q.7 The formula used to calculate mass moment of inertia (I G) of a
torsional pendulum or Flywheel about the axis through Centre of gravity?
0.5 m r2
m r2
0.25 m r2 None of the options
Q.8 The natural frequency of vibrations of a torsional pendulum with the
following dimensions. Length of rod(L) = 1 m, Diameter of rod(d) =5
mm, Diameter of the rotor(D)= 0.2 m Mass of the rotor (M) = 2 Kg.
Modulus of rigidity for the material of the rod may be assumed to be 0.83
x 10 11 N/ m2 will be
4.72 Hz.
6.29 Hz 3.59 Hz.
5.54 Hz.
Q.9 A six cylinder in-line engine directly drives a propeller through a
coupling. The degree of freedom of this torsional system will be
3
8
6
4
Q.10 In a Semi-definite or Degenerate system one of the natural
frequencies will be
Infinity
Hz
Two Hz
Unity Hz Zero Hz
MCQ MDL 6
Q.1 In a torsional vibrating system when the rotor A, the rotor B, the
rotor C, rotate in the directions shown in Fig. The number of nodes
formed will be
Zero nodes One node Two nodes Three nodes
Q.2 For a ship “ The critical speed speed range” or “The barred speed
speed range “ is due to the following phenomenon
Beat phenomenon
Resonance phenomenon
Interference phenomenon
None of the options
Q.3 A torsional vibrating system carrying four rotors and has a degree
Of freedom(DOF) of four. It’s four natural frequencies are fn0= 0 Hz,
fn1= 32 Hz, fn2= 143 Hz, fn3= 287 Hz. When this system was vibrating,
two nodes were observed to be formed. Then in that case the might be
vibrating at
0 Hz
32 Hz
143 Hz
287 Hz
Q.4 In case of a torsional vibrating system, the internal induced stresses
Will be
Tensile stresses
Bending stresses
Torsional shear stresses Compressive stresses
Q.5 In torsional vibrations at node points
Torsional shear stresses are maximum
Torsional shear stresses are minimum
No torsional shear stresses are induced
None of the options
Q.6 In the torsional vibrating system when the rotor A, the rotor B, the
rotor C, rotate in the direction as shown in Fig. The number of nodes
formed will be
Zero nodes One node Two nodes Three nodes
Q.7 The formula used to calculate polar moment of inertia of shaft
carrying torsional pendulum or Flywheel about the axis through centre of
gravity?
(Pi X d4)/32
(Pi X d4)/64
(Pi X d6)/32
None of the options
Q.8 The natural frequency of vibrations of a torsional pendulum with the
following dimensions. Length of rod(L) = 1.2 m, Diameter of rod(d) =8
mm, Diameter of the rotor(D)= 0.4 m Mass of the rotor (M) = 2.5 Kg.
Modulus of rigidity for the material of the rod may be assumed to be 0.83
x 1011 N/ m2 will be
4.72 Hz.
6.29 Hz 3.75 Hz.
5.54 Hz.
Q.9 A four cylinder in-line engine directly drives a propeller through a
coupling. The degree of freedom of this torsional system will be
3
8
6
4
Q.10 In torsional vibrations as frequency increases, the number of nodes
will
Increase
Decrease Remain unchanged
None of the options
MCQ MDL 7
TRANSVERSE VIB
Q.1 Vibrations of a shaft due to the weights of various components like
Gears, pulleys, sprockets etc will be
Longitudinal vibrations Torsional vibrations
Transverse vibrations
Axial vibrations
Q.2 In case of a Transverse vibrating system, the internal particles of
the system would be found to be moving in
Along the lines parallel to the axis of the vibrating system
Along the lines Perpendicular to the axis of the vibrating system
In concentric circles about the axis of the vibrating system
None of the options
Q.3 Natural Frequency of Free Transverse Vibrations For a Shaft
Subjected to a Number of Point Loads is determined by
Dunkerly’s method
Newton’s method
D’Alembert’s method
None of the options
Q.4 In case of a Transverse vibrating system, the internal induced
stresses Will be
Tensile stresses
Bending stresses
Torsional shear stresses Compressive stresses
Q.5 The static deflection of a Cantilever beam with end load is given by
𝐩 𝐥𝟑
𝟑𝐄𝐈
None of the options
𝐩 𝐥𝟒
𝐩 𝐥𝟐
𝟑𝐄𝐈
𝟔𝐄𝐈
Q.6 The static deflection of a Simply supported beam with a central load
is given by
𝐰 𝐥𝟑
𝟒𝟖𝐄𝐈
𝐰 𝐥𝟐
𝐰 𝐥𝟒
𝟐𝟒𝐄𝐈
𝟒𝟖𝐄𝐈
None of the options
Q.7 The static deflection of a Simply supported beam with a off-centre
load is given by
𝐰 𝒍𝟐𝟏 𝒍𝟐𝟐
𝟑𝐄𝐈𝐥
𝐰 𝒍𝟑𝟏 𝒍𝟐𝟐
𝐰 𝒍𝟐𝟏 𝒍𝟑𝟐
𝟒𝟖𝐄𝐈𝐥
𝟒𝟖𝐄𝐈𝐥
None of the options
Q.8 The static deflection of a Cantilever beam with non-end load is given
by
𝐰 𝒍𝟑𝟏
𝟑𝐄𝐈
𝐰 𝒍𝟐𝟏
𝐰 𝒍𝟒𝟏
𝟑𝐄𝐈
𝟑𝐄𝐈
None of the options
Q.9 The static deflection of a Fixed-fixed beam with a central load is
given by
𝐰 𝐥𝟑
𝟏𝟗𝟐𝐄𝐈
𝐩 𝐥𝟐
𝐩 𝐥𝟒
𝟏𝟗𝟐𝐄𝐈
𝟏𝟗𝟐𝐄𝐈
None of the options
Q.10 The static deflection of a beam due to uniformly distributed load is
given by
𝟓𝐖 𝐥𝟒
𝟑𝟖𝟒𝐄𝐈
𝟓𝐖 𝐥𝟑
𝟓𝐖 𝐥𝟐
𝟑𝟖𝟒𝐄𝐈
𝟑𝟖𝟒𝐄𝐈
None of the options
MCQ MDL 8
WHIRLING-SHAFT
Q.1 The critical speed or whirling speed of a shaft is due to
Centrifugal force Centripetal force Unbalance force
None of the options
Q.2 The critical speed or whirling speed of a shaft is the speed at which
The shaft starts vibrating in a direction
Parallel to the axis of the vibrating system
Perpendicular to the axis of the vibrating system
In concentric circles about the axis of the vibrating system
None of the options
Q.3 The deflection of the shaft at whirling speed is given by
𝛚 𝟐
) 𝐞
𝛚𝐧
𝐲=
𝛚 𝟐
𝟏−( )
𝛚𝐧
(
𝛚 𝟑
( ) 𝐞
𝛚𝐧
𝐲=
𝛚 𝟐
𝟏−( )
𝛚𝐧
𝛚 𝟒
( ) 𝐞
𝛚𝐧
𝐲=
𝛚 𝟐
𝟏−( )
𝛚𝐧
None of the options
Q.4 The speed of a shaft at which the deflection of the shaft tends to
Infinity is known as
Whirling speed
Peak speed
Maximum speed None of the options
Q.5 When the shaft speed is less than the critical speed, the deflection ‘y’
will be
Positive
Nagative
Neutral None of the options
Q.6 When the shaft speed is equal to the critical speed, the deflection ‘y’
will
Increase Decrease
Tend to Infinity None of the options
Q.7 When the shaft speed is more than the critical speed, the deflection
‘y’
will be
Positive
Nagative
Neutral None of the options
Q.8 The critical speed or whirling speed of a shaft is
Inversely proportional to eccentricity
Directly proportional to eccentricity
Independent of eccentricity
None of the options
Q.9 Centrifugal force that makes the shaft vibrate violently is
Inversely proportional to eccentricity
Directly proportional to eccentricity
Independent of eccentricity
None of the options
Q.10 The deflection of the shaft at whirling speed is
Inversely proportional to frequency ratio
Directly proportional to frequency ratio
Independent of frequency ratio
None of the options