Forced Vibration MCQ Quiz in తెలుగు - Objective Question with Answer for Forced Vibration - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Last updated on Apr 16, 2025

పొందండి Forced Vibration సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి Forced Vibration MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest Forced Vibration MCQ Objective Questions

Top Forced Vibration MCQ Objective Questions

Forced Vibration Question 1:

In a forced vibration system, for which value of frequency ratio (ωfn), the transmissibility is same for all the values of damping factors

  1. 1
  2. 2
  3. √2
  4. 1/2

Answer (Detailed Solution Below)

Option 3 : √2

Forced Vibration Question 1 Detailed Solution

Concept:

In vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

SSC CPU 60

T=FTFO=1+(2ζωωn)2[1(ωωn)2+(2ζωωn)2]

  • When ω/ωn = 0 ⇒ TR = 1, (independent of ζ)
  • When ω/ωn = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
  • When frequency ratio ω/ωn = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
  • When frequency ratio ω/ωn < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
  • When frequency ratio ω/ωn > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ωn > √2. Here the force transmitted to the foundation increases as the damping is increased.

Forced Vibration Question 2:

In vibration isolation system, if ω/ωn = 2 then transmissibility will be

  1. > 1
  2. = 1
  3. < 1
  4. zero

Answer (Detailed Solution Below)

Option 2 : = 1

Forced Vibration Question 2 Detailed Solution

Explanation:

Transmissibility ratio (TR):

In a vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

SSC CPU 60

TR=FTFO=1+(2ζωωn)2[1(ωωn)2+(2ζωωn)2]

  • When ω/ωn = 0 ⇒ TR = 1, (independent of ζ)
  • When ω/ωn = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
  • When frequency ratio ω/ωn = 2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
  • When frequency ratio ω/ωn < 2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
  • When frequency ratio ω/ωn > 2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ωn > 2. Here the force transmitted to the foundation increases as the damping is increased.

Forced Vibration Question 3:

A machine of 250 kg mass is supported on springs of total stiffness 100 kN/m. Machine has an unbalanced rotating force of 350 N at speed of 3600 rpm. Assuming a damping factor of 0.15, the value of transmissibility ratio is:

  1. 0.0531
  2. 0.9922
  3. 0.0162
  4. 0.0028

Answer (Detailed Solution Below)

Option 3 : 0.0162

Forced Vibration Question 3 Detailed Solution

ω=2π×360060=377rad/s

Natural frequency, ωn=km=100×1000250=20 rad/s

Now, r=ωωn=37720=18.85

Transmissibility ratio (TR)=1+(2ξr)2(1r2)2+(2ξr)2

=1+(2×0.15×18.85)2[1(18.85)2]2+(2×0.15×18.85)2

= 0.0162

Forced Vibration Question 4:

A column AB 10 m long and 30 kN weight is hinged at A and is supported by a spring of stiffness K as shown in the figure. Which of the following statements are true?

  1. If k = 1.5 kN / m, Column is in neutral equilibrium
  2. If k > 1.5 kN / m, Column is in stable equilibrium
  3. If k < 1.5 kN / m, Column is in unstable equilibrium
  4. Cannot be determined.

Answer (Detailed Solution Below)

Option :

Forced Vibration Question 4 Detailed Solution

Concept:

Stable equilibrium: If a small displacement is given to the body, it will return to its original position and will be in equilibrium

Unstable equilibrium: If a small displacement is given to the body, it will continue to move away from its original position and will not be in equilibrium

Neutral equilibrium: If a small displacement is given to the body, it will occupy a new position in equilibrium.

Calculation:

Draw FBD

F2 Sumit Madhu 19.08.20 D5

Apply moment about A,

Given length (L) = 10 m

Weight (w) = 30 KN

F×L=W×x2(1)

Kx×10=W×x2

K=1.5kNm (Neutral equilibrium)

If K > 1.5 kN/m, the bar will restore its original position

If K < 1.5 kN/m, the bar will never restore its original position

Forced Vibration Question 5:

A damped mass spring system has mass of 10 kg, spring stiffness of 4000 N/m and damping coefficient of 40 Ns/m. The amplitude of forcing function, F0 is 60 N and the forcing frequency is 40 rad/sec. Determine the transmissibility of the system. (round up to two decimal places)

Answer (Detailed Solution Below) 0.31 - 0.38

Forced Vibration Question 5 Detailed Solution

Concept:

FT is the force transmitted to the foundation. The disturbing force is F. The ratio of FT to F is called transmissibility.

T.R=FTF=1+(2ξωωn)2(1(ωωn)2)2+(2ξωωn)2

The steady state amplitude for the system:

A=Fk(1(ωωn)2)2+(2ξωωn)2

Calculation

Given: m = 10 kg, k = 4000 N/m, C = 40 Ns/m, F = 60 N

ω = 40 rad/sec

ωn=km=400010=20rad/sec

Frequency Ratio:

r=ωωn=4020=2

The critical damping coefficient:

cc=2mωn=2mk=210×4000=400N.s/m

The damping factor, ξ:

ξ=CCc=40400=0.1

Transmissibility:

T.R=FTF=1+(2ξωωn)2(1(ωωn)2)2+(2ξωωn)2

T=1+(2×0.1×2)2(122)2+(2×0.1×2)=1.077033.1875=0.3379

Forced Vibration Question 6:

Given a machine that is connected to the ground whose law of motion is given below. What is the maximum force in N that can get transmitted to the ground?

3x ¨+14x˙+72x=200cos(9t)

Answer (Detailed Solution Below) 135 - 138

Forced Vibration Question 6 Detailed Solution

Concept:

Motion of Forced damped vibration:

mx ¨+cx˙+kx=Fcos(ωt) ------(1)

ωn=Keqm

Damping factor: ξ=ccc=c2km

T=FTFun=1+(2ξωωn)2(1(ωωn)2)2+(2ξωωn)2 

Calculation:

Given:

3x ¨+14x˙+72x=200cos(9t)

On comparing the above equation with equation 1

m = 3 kg, c = 14 Ns/m, K = 72 N/m, ω = 9 rad/s, F = 200 N

ωn=Keqm=723=4.9 rad/s

ξ=c2km=142×72×3=0.4763

ωωn=94.9=1.837

T=FTFun=1+(2ξωωn)2(1(ωωn)2)2+(2ξωωn)2 

T=1+(2×0.4763×1.837)2(1(1.837)2)2+(2×0.4763×1.837)2 =2.0152.95=0.683

FT=T×Fun=0.683×200=136.6 N

Forced Vibration Question 7:

If the damping coefficient is 10 Ns/m, mass of the body is 5 kg and the angular velocity is 240 rad/min, then damping ratio is:

  1. 0.250
  2. 0.004
  3. 0.050
  4. 0.008

Answer (Detailed Solution Below)

Option 1 : 0.250

Forced Vibration Question 7 Detailed Solution

Concept:

Damping ratio:

The ratio of the actual damping coefficient (c) to the critical damping coefficient (cc) is known as damping factor or damping ratio.

 

ζ=CCc=C2km

Also, 2ξωn=Cm

F2 S.S-D.K 05.09.2019 D1

 

  • Overdamped System: ζ > 1
  • Underdamped: ζ < 1
  • Critical Damping: ζ = 1: The displacement will be approaching to zero in the shortest possible time. The system does not undergo a vibratory motion. 

Calculation:

Given:

C = 10 Ns/m, m = 5 kg, ωn =  240 rad/min = 4 rad/sec

Therefore, for damping ratio, 

2ξωn=Cm

2ξ×4=105

ξ = 0.25

Forced Vibration Question 8:

In a vibration isolation which of the following condition states that the phase difference between the unbalanced force and transmitted force is 180˚?

  1. ωn/ω = 1
  2. ωn/ω < 1
  3. ωn/ω > √2
  4. ωn/ω < √2

Answer (Detailed Solution Below)

Option 2 : ωn/ω < 1

Forced Vibration Question 8 Detailed Solution

For ω/ωn > 1 the phase will be 180˚ as the value 1- (ω/ωn)2 will become negative which means out of phase.

Phase angle:

ϕ=tan1(2ξωωn1(ωωn)2)

Phase angle varies from zero at low frequencies to 180° at very high frequencies. It changes very rapidly near the resonance and is 90° at resonance irrespective of damping.

The phase angle between 0 to 90° for all the values of ω/ωn = 0 to 1 and between 90° to 180° for all the values of ω/ωn > 0

In absence of any damping, phase angle suddenly changes from zero to 180° at resonance.

The phase angle φ is zerorrg for all the values of damping ratio at ω/ωn = 0

Forced Vibration Question 9:

A forced damped vibration system is represented as 5x¨+20x˙+80x=8cos(4t). What is the magnification factor?

  1. 1
  2. 1.5
  3. 0.5
  4. 2

Answer (Detailed Solution Below)

Option 1 : 1

Forced Vibration Question 9 Detailed Solution

Concept:

MF=1(1r2)2+(2ξr)2

Calculation:

Comparing with mx¨+cx˙+kx=F0cosωt

m = 5 kg, c = 20 Ns/m, k = 80 N/m, ω = 4 rad/s

ωn=km=805=4rad/s

ξ=C2km=20280×5=0.5

MF=1(1(ωωn)2)2+(2ξ(ωωn))2

MF=12ξ

∴ MF = 1

Forced Vibration Question 10:

When the frequency of the vibrating system becomes equal to the natural frequency of the system, this type of vibration is known as:

  1. Magnification factor
  2. Random vibration
  3. Forced vibration
  4. Resonance

Answer (Detailed Solution Below)

Option 4 : Resonance

Forced Vibration Question 10 Detailed Solution

Explanation:

Resonance:

  • The phenomenon of increase in amplitude when the frequency of driving force is close to the natural frequency of the oscillator is called resonance.
  • Therefore for the resonance of the forced oscillation of the spring-mass system,

⇒ ω = ωn

Where ωn = natural frequency, and ω = driving frequency

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