Theory of Machines Miscellaneous
- The differential equation governing the vibrating system is
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This is the differential equation governing the above vibrating system.Correct Option: C
This is the differential equation governing the above vibrating system.
- In a spring-mass system, the mass is 0.1 kg and the stiffness of the spring is 1 kN/m. By introducing a damper, the frequency of oscillation is found to be 90% of the original value. What is the damping coefficient of the damper?
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8.7 Ns/m
Correct Option: C
8.7 Ns/m
- There are four samples P, Q, R and S with natural frequencies 64,96,128 and 256 Hz, respectively. They are mounted on test setups for conducting vibration experiments. If a loud pure note of frequency 144 Hz is produced by some instrument, which of the samples will show the most perceptible induced vibration?
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R produces vibration close to natural frequency 144 Hz.
30. d = 0.9 fn
⇒ ωd = 0.9 ωn
⇒ √ωn2 - α2 = 0.9 ωn
Squaring ωn2 - α2 0.81ωn2
⇒ 0.19 ωn2 = α2⇒ 0.19 × s = c2 m (2m)102
⇒ c2 = 0.76 sm
⇒ c = √0.76 × 1 × 103 × 0.1
= 8.717 Ns/mCorrect Option: C
R produces vibration close to natural frequency 144 Hz.
30. d = 0.9 fn
⇒ ωd = 0.9 ωn
⇒ √ωn2 - α2 = 0.9 ωn
Squaring ωn2 - α2 0.81ωn2
⇒ 0.19 ωn2 = α2⇒ 0.19 × s = c2 m (2m)102
⇒ c2 = 0.76 sm
⇒ c = √0.76 × 1 × 103 × 0.1
= 8.717 Ns/m
- A mass m, of 20 kg is attached to the free end of a steel cantilever beam of length 1000 mm having a cross-section of 25 × 25 mm. Assume the mass of the cantilever to be negligible and Esteel = 200 GPa. If the lateral vibration of this system is critically damped using a viscous damper, the damping constant of the damper is
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Given length of cantilever beam,
l = 1000 mm = 1 m
Moment of inertia of the shaft,I = 1 bd3 = 25 × (25)3 12 12
= 3.25 × 10-8 m4
Esteel = 200 × 109 Pa.
Mass, M = 20 kg
∴ Weight W = 20 × 9.81 = 196.2
Now, deflection of free end,δ = Wl3 = 196.2 × (1)3 = 0.01 m 3EI 3 × 200 × 109 × 3.25 × 10-8 
As the system is critically damped,
Cc 
2 = ωn2 2m
Critical damping constant,
Cc = 2mωn = 2 × 20 × 31.22
= 1248.99 N-s/m2.Correct Option: A
Given length of cantilever beam,
l = 1000 mm = 1 m
Moment of inertia of the shaft,I = 1 bd3 = 25 × (25)3 12 12
= 3.25 × 10-8 m4
Esteel = 200 × 109 Pa.
Mass, M = 20 kg
∴ Weight W = 20 × 9.81 = 196.2
Now, deflection of free end,δ = Wl3 = 196.2 × (1)3 = 0.01 m 3EI 3 × 200 × 109 × 3.25 × 10-8
As the system is critically damped,Cc 2 = ωn2 2m
Critical damping constant,
Cc = 2mωn = 2 × 20 × 31.22
= 1248.99 N-s/m2.
Direction: A uniform rigid slender bar of mass 10 kg, hinged at the left end is suspended with the help of spring and damper arrangement as shown in the figure where K= 2 kN/m, C = 500 Ns/m and the stiffness of the torsional spring kθ is 1 kN /m/rad. Ignore the hinge dimensions.
- The damping coefficient in the vibration equation is given by
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For static equilibrium kt . θs + kt . Theta;s L2 = mgl2
For additional deflection θ, equation will be.
where, I = ml2/3
∴ Un-damped frequency,
= 42.426
= 42.43 rad/sec.
and, Equivalent damping coefficient
= Cl12 Nm/rad
= 500 × (0.04)2 = 80 Nms/radCorrect Option: C
For static equilibrium kt . θs + kt . Theta;s L2 = mgl2
For additional deflection θ, equation will be.
where, I = ml2/3
∴ Un-damped frequency,
= 42.426
= 42.43 rad/sec.
and, Equivalent damping coefficient
= Cl12 Nm/rad
= 500 × (0.04)2 = 80 Nms/rad
