Theory of Machines Miscellaneous
- A single degree of freedom spring mass system with viscous damping has a spring constant of 10 kN/m. The system is excited by a sinusoidal force of amplitude 100 N. If the damping factor (ratio) is 0.25, the amplitude of steady state oscillation at resonance is _______ mm.
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k = 10 kN/m
F0 = 100 N
ξ = 0.25
ω = 1 at resonance ωn X = F0 = 100 = 20 mm kξ 2 × 100 × .25 × 10³ Correct Option: B
k = 10 kN/m
F0 = 100 N
ξ = 0.25ω = 1 at resonance ωn X = F0 = 100 = 20 mm kξ 2 × 100 × .25 × 10³
- A single-degree-freedom spring-mass system is subjected to a sinusoidal force of 10 N amplitude and frequency ω along the axis of the spring. The stiffness of the spring is 150 N/m, damping factor is 0.2 and the undamped natural frequency is 10 ω. At steady state, the amplitude of vibration (in m) is approximately
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Amplitude of vibration,
Correct Option: B
Amplitude of vibration,
- A precision instrument package (m = 1 kg) needs to be mounted on a surface vibrating at 60 Hz. It is desired that only 5% of the base surface vibration amplitude be transmitted to the instrument. Assuming that the isolation is designed with its natural frequency significantly lesser than 60 Hz, so that the effect of damping may be ignored. The stiffness (in N/m) of the required mounting pad is ________.
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ω = 2πN = 2 × π × 60 = 376.99 rad/second
.05 = 1 1 - (ω/ωn)² ⇒ -20 = 1 - 
ω 
² ⇒ ω ² = 21 ωn ωn
ω2 = 21ω²n
ωn = 82.266 rad/s
ωn = √k/m = 82.266
1 = 6767.6005 N/mCorrect Option: B
ω = 2πN = 2 × π × 60 = 376.99 rad/second
.05 = 1 1 - (ω/ωn)² ⇒ -20 = 1 - ω ² ⇒ ω ² = 21 ωn ωn
ω2 = 21ω²n
ωn = 82.266 rad/s
ωn = √k/m = 82.266
1 = 6767.6005 N/m
- Torque and angular speed data over one cycle for a shaft carrying a flywheel are shown in the figures. The moment of inertia (in kgm²) of the flywheel is _______ is.
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I =?
E = 3000 × 
π - π 
- 1500[2π - π] = 0 2
Tmean × 2π = E
Tmean = 0
ωmin = 10 rad/s at θ = π/2
ωmax = 20 rad/s at θ = πCs = ωmax - ωmin = 20 - 10 ωmax + ωmin 20 + 10 2 2 = 10 × 2 = 0.67 30 ΔE = 3000 × π = 1500π = Iω²CS 2
1500π = I × ω²mean × 0.67
= 1500 Δ = I × 15² × 0.67
I = 31.26 kg-m²Correct Option: D
I =?
E = 3000 × π - π - 1500[2π - π] = 0 2
Tmean × 2π = E
Tmean = 0
ωmin = 10 rad/s at θ = π/2
ωmax = 20 rad/s at θ = πCs = ωmax - ωmin = 20 - 10 ωmax + ωmin 20 + 10 2 2 = 10 × 2 = 0.67 30 ΔE = 3000 × π = 1500π = Iω²CS 2
1500π = I × ω²mean × 0.67
= 1500 Δ = I × 15² × 0.67
I = 31.26 kg-m²
- Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg.m²) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is __________.
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ΔE = 2600 J
Naverage = 200 rpm
I =?
CS = ± 0.5%= 2 × 0.5 = 0.1 100
ΔE = Iω²CS2600 = I × 2π × 200 ² × 0.01 60
I = 592.73 kg-m²Correct Option: D
ΔE = 2600 J
Naverage = 200 rpm
I =?
CS = ± 0.5%= 2 × 0.5 = 0.1 100
ΔE = Iω²CS2600 = I × 2π × 200 ² × 0.01 60
I = 592.73 kg-m²
