In centrifugal pendulum absorber , the natural frequency in cycle per second can be given by
A) Fn =N √(R/L)
B) Fn =1/N √(R/L)
C) Fn =N/2 √(R/L)
D) Fn =N2 √(R/L)

1 Answer

Answer :

A) Fn =N √(R/L)

Related questions

Description : . In centrifugal pendulum absorber , the natural frequency in cycle per second can be given by A Fn =N √(R/L) B Fn =1/N √(R/L) C Fn =N/2 √(R/L) D Fn =N2 √(R/L)

Last Answer : A Fn =N √(R/L)

Description : Frequency of centrifugal pendulum absorber is always proportional to A) Oscillating motion B) Transfer motion C) Speed of rotating body D) All of the above

Last Answer : C) Speed of rotating body

Description : The natural frequency of torsional vibration is given by A) ωn = (-kt)/I B) ωn = kt/I C) ωn = √(kt/I) D) ωn = √(2&kt/I)

Last Answer : C) ωn = √(kt/I)

Description : Centrifugal absorber is used to reduce A) Centrifugal force in rotating system B) Torsional vibration of rotating system C) Vibration in linear system D) Transverse vibrations

Last Answer : B) Torsional vibration of rotating system

Description : Which of the following condition should be satisfied in the design of a vibration absorber ? A) Natural frequency of the auxiliary system should be equal to the natural frequency of the main ... D) Natural frequency of the auxiliary system should be twice natural frequency of the main system

Last Answer : A) Natural frequency of the auxiliary system should be equal to the natural frequency of the main system

Description : Natural frequency of simple pendulum is proportional to A. Length B. Acceleration due to gravity C. Both D. None

Last Answer : B. Acceleration due to gravity

Description : The two resonant frequency ratio (ω/ ω2) in a dynamic vibration absorber system for a mass ratio 0.2 are given by A) 0 ; 1.0 B) 0.801 ; 1.248 C) 0.458 ; 1.124 D) 0.642 ; 1.558

Last Answer : B) 0.801 ; 1.248

Description : What is a conical pendulum? Show that its time period is given by 2π √(l cos θ)/g, where l is the length of the string,

Last Answer : What is a conical pendulum? Show that its time period is given by 2π\(\sqrt{\frac{l\,cos\, ... the vertical and g is the acceleration due to gravity.

Description : Consider the steady-state absolute amplitude equation shown below, if ω / ω n = √2 then amplitude ratio (X/Y) =? (X/Y) = √{1 + [ 2ξ (ω/ω n )] 2 } / √{[1 – (ω/ω n ) 2 ] 2 + {2ξ (ω/ω n ) 2 } A. 0 B. 1 C. less than 1 D. greater than 1

Last Answer : B. 1

Description : for the system shown below K1=20N/m K1=10N/m K1=20N/m K1=50N/mFind W such that the natural frequency of the system will be 1.592 cycles per second a)0.125kg b)0.25kg c)0.5kg 4)4kg

Last Answer : b)0.25kg

Description : A simple pendulum is found to vibrate at a frequency of 0.5Hz in vacuum and0.45 Hz in a viscous fluid medium. Find the damping factor. 0.5122 (B) 0.9272 (C) 0.4359 (D) 0.2568

Last Answer : (C) 0.4359

Description : A simple pendulum is found to vibrate at a frequency of 0.5Hz in vacuum and 0.45Hz in a viscous fluid medium. Find the damping factor. A 0.5122 B 0.9237 C 0.4359 D 0.2568

Last Answer : C 0.4359

Description : Critical speed of rotation, N (in rps - rotation per second) of a trammel is equal to (where, g = acceleration due to gravity = 9.81 m/sec 2 and, r = radius of trammel, metre.) (A) (1/2π). √(g/r) (B) (1/π). √(g/r) (C) ½ √(g/r) (D) 2π. √(g/r)

Last Answer : (A) (1/2π). √(g/r)

Description : A shaft of length l carries two discs at its two ends. The lowest torsional frequency is ω n . If the shaft length is doubled, then the lowest torsional frequency becomes A ω n /2 B ω n /√2 C √2ω n D 2ω n

Last Answer : B ω n /√2

Description : The number of cycles described in one second is known as _______. A period of vibration B cycle C frequency D all of the above

Last Answer : C frequency

Description : The number of cycles described in one second is known as _______. C ( A ) period of vibration ( B ) cycle ( C ) frequency ( D ) all of the above

Last Answer : C ) frequency

Description : Determine natural frequency of a system, which has equivalent spring stiffness of 30000 N/m and mass of 20 kg? A 12.32 Hz B 4.10 Hz C 6.16 HzD None of the above

Last Answer : C 6.16 Hz

Description : A 1 kg mass is suspended by a spring having a stiffness of 0.4 N/mm. Determine the natural frequency. A 20 rad/sec B 30 rad/sec C 20 Hz D 30 Hz

Last Answer : B 30 rad/sec

Description : Determine natural frequency of a system, which has equivalent spring stiffness of 43200 N/m and mass of 12 kg. A 40.22 rad/sec B 40 Hz C 60 Hz D 60 rad/sec

Last Answer : D 60 rad/sec

Description : Determine natural frequency of a system, which has equivalent spring stiffness of 30000 N/m and mass of 20 kg? A. 12.32 Hz B. 4.10 Hz C. 6.16 Hz D. None of the above

Last Answer : C. 6.16 Hz

Description : Calculate coefficient of viscous damper, if the system is critically damped. Consider the following data: 1. Mass of spring mass damper system = 350 kg 2. Static deflection = 2 x 10 -3 m 3. Natural frequency of the system = 60 rad/sec ... /m B. 80 x 10 3 N-s/m C. 42 x 10 3 N-s/m D. None of the above

Last Answer : C. 42 x 10 3 N-s/m

Description : Calculate 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 A. 3.99 rad/sec B. 2.13 rad/sec C. 4.12 rad/sec D. 3.24 rad/sec

Last Answer : D. 3.24 rad/sec

Description : A system has a mass of 0.5 kg and spring stiffness of 2452 N/m. Find the natural frequency of the system. A. 5.14 Hz B. 9.14 Hz C. 11.14 Hz D. 28.14 Hz

Last Answer : C. 11.14 Hz

Description : Determine natural frequency of a system, which has equivalent spring stiffness of 30000 N/m and mass of 20 kg? A) 12.32 Hz B) 4.10 Hz C) 6.16 Hz D) None of the above

Last Answer : C) 6.16 Hz

Description : Determine natural frequency of a system, which has equivalent spring stiffness of 30000 N/m and mass of 20 kg? C ( A )12.32 Hz (B) 4.10 Hz ( C )6.16 Hz (D)None of the above

Last Answer : ( C )6.16 Hz

Description : In spring mass experiment, the natural frequency of 10 kg mass was found to be 12 rad/sec. the stiffness of the spring is A. 800 N/m B. 1200 N/m C. 1440 N/m D. 2000 N/m

Last Answer : C. 1440 N/m

Description : A 10 Kg mass suspended by spring of stiffness 1000 N/m. the natural frequency of the system after giving excitation will be A. 0 Hz B. 1.59 Hz C. 2 Hz D. 15.9 Hz

Last Answer : B. 1.59 Hz

Description : A vehicle suspension system consists of a spring and a damper. The stiffness of the spring is 3.6 kN/m and the damping constant of the damper is 400 Ns/m. If the mass is 50 kg, then the damping factor (d ) and damped natural ... , are a) 0.471 and 1.19 Hz b) 0.471 and 7.48 Hz c) 0.666 and 1.35 Hz

Last Answer : a) 0.471 and 1.19 Hz

Description : The natural frequency of a spring-mass system on earth is ω n . The natural frequency of this system on the moon (g moon = g earth /6) is a) ω n b) 0.408ω n c) 0.204ω n d) 0.167ω n

Last Answer : a) ω n

Description : Determine natural frequency of a system, which has equivalent spring stiffness of 30000 N/m mass of 20 kg? a. 12.32 Hz b. 4.10 Hz c. 6.16 Hz d. None of the above

Last Answer : c. 6.16 Hz

Description : Calculate coefficient of viscous damper, if the system is critically damped. Consider the following data: 1. Mass of spring mass damper system = 350 kg 2. Static deflection = 2 x 10 -3 m 3. Natural frequency of the system = 60 rad/sec ... /m b. 80 x 10 3 N-s/m c. 42 x 10 3 N-s/m d. None of the above

Last Answer : c. 42 x 10 3 N-s/m

Description : A spring-mass system has a natural frequency of 10 rad/sec. When the spring constant is reduced by 800 N/m, the frequency is altered by 45 percent. Find the mass and spring constant of the original system. a)11.47kg and 1147.95N/m b)8.95kg and 895.25N/m c) 7.265kg and 726.5N/m d)None

Last Answer : a)11.47kg and 1147.95N/m

Description : A system has a mass of 0.5 kg and spring stiffness of 2452 N/m. Find the natural frequency of the system. (A) 5.14 Hz (B) 9.14 Hz (C) 11.14 Hz (D) 28.14 Hz

Last Answer : (C) 11.14 Hz

Description : A vehicle suspension system consists of a spring and a damper. The stiffness of the spring is 3.6 kN/m and the damping constant of the damper is 400 Ns/m. If the mass is 50 kg, then the damping factor (d ) and damped natural ... .19 Hz b) 0.471 and 7.48 Hz c) 0.666 and 1.35 Hz d) 0.666 and 8.50 Hz

Last Answer : a) 0.471 and 1.19 Hz

Description : The natural frequency of a spring-mass system on earth is ω n . The natural frequency of this system on the moon (g moon = g earth /6) is a) ω n b) 0.408ω n c) 0.204ω n d) 0.167ω n

Last Answer : a) ω n

Description : In a dynamic vibration absorber system, under tuned conditions which of the following relation holds good? A K 1 K 2 =M 1 M 2 B K 1 M 2 =M 1 K 2 C K 1 M 1 = K 2 M 2 D none of the mentioned

Last Answer : B K 1 M 2 =M 1 K 2

Description : Dynamic vibration absorber is suitable for A varying speed machines B constant speed machines C Zero speed range machines D None of the mentioned

Last Answer : B constant speed machines

Description : Dynamic vibration absorber is suitable for A) Constant speed machine B) Varying speed machine C) zero speed machine D) None of the above

Last Answer : A) Constant speed machine

Description : When ≠ , the absorber is known as A. Untuned vibration absorber B. Tuned vibration absorber C. Both of above D. None of these

Last Answer : A. Untuned vibration absorber

Description : When = , the absorber is known as A. Untuned vibration absorber B. Tuned vibration absorber C. Both of above D. None of these

Last Answer : B. Tuned vibration absorber

Description : Dynamic vibration absorber is suitable for A. Constant speed machines B. Varying speed machines C. Zero speed range machines D. None of the above

Last Answer : A. Constant speed machines

Description : Frequency of vibrations is usually expressed in ( A ) number of cycles per hour ( C ) number of cycles per second C ( B ) number of cycles per minute ( D ) None of these

Last Answer : ( C ) number of cycles per second

Description : When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as * 1 point (A) simple pendulum (B) torsional pendulum (C) compound pendulum (D) second’s pendulum

Last Answer : (C) compound pendulum

Description : When the torsional pendulum vibrating the observed amplitudes on the same side of neutral axis for successive cycles are found to decay 50% of the initial value determine logarithmic decrement. A 0.065 B 0.006 C 0.693 D 0.5

Last Answer : C 0.693

Description : When the torsional pendulum vibrating, the observed amplitudes on the same side of the neutral axis for successive cycles are found to decay 50% of the initial value determine logarithmic decrement. A 0.065 B 0.006 C 0.693 D 0.5

Last Answer : C 0.693

Description : When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as A. simple pendulum B. torsional pendulum C. compound pendulum D. second’s pendulum

Last Answer : C. compound pendulum

Description : Who discovered the concept of pendulum? A. Newton B. Einstein C. Galileo D. None

Last Answer : C. Galileo

Description : The number of degrees of freedom of a simple pendulum is: (a) 0 (b) 1 (c) 2

Last Answer : (b) 1

Description : The time period of a simple pendulum does not depend upon the mass of the body suspended at the free end of the string. This statement is known as ___________ . (A) law of gravity (B) law of mass (C) law of isochronism (D) law of length

Last Answer : (B) law of mass

Description : When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the force of gravity, the body is known as (A) simple pendulum B) torsional pendulum (C) compound pendulum (D) second’s pendulum

Last Answer : (C) compound pendulum