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This section includes 8 Mcqs, each offering curated multiple-choice questions to sharpen your Machine Dynamics knowledge and support exam preparation. Choose a topic below to get started.
1. |
Frequency is independent of the no. of nodes. |
A. | True |
B. | False |
Answer» C. | |
2. |
A single cylinder oil engine works with a three rotor system, the shaft length is 2.5m and 70mm in diameter, the middle rotor is at a distance 1.5m from one end. Calculate the free torsional vibration frequency for a two node system in Hz if the mass moment of inertia of rotors in Kg-m2 are: 0.15, 0.3 and 0.09. C=84 kN/mm2 |
A. | 257 |
B. | 281 |
C. | 197 |
D. | 277 |
Answer» E. | |
3. |
A single cylinder oil engine works with a three rotor system, the shaft length is 2.5m and 70mm in diameter, the middle rotor is at a distance 1.5m from one end. Calculate the free torsional vibration frequency for a single node system in Hz if the mass moment of inertia of rotors in Kg-m2 are: 0.15, 0.3 and 0.09. C=84 kN/mm2 |
A. | 171 |
B. | 181 |
C. | 191 |
D. | 201 |
Answer» B. 181 | |
4. |
For a vibration system having different shaft parameters, calculate which of the following cannot be the diameter of the equivalent shaft if the diameters of shafts in m are: 0.05, 0.06, 0.07. |
A. | 0.05 |
B. | 0.06 |
C. | 0.07 |
D. | 0.08 |
Answer» E. | |
5. |
From the following data, calculate the location of node from the left end of shaft (l1).l1=0.6m, l2=0.5m, l3=0.4md1=0.095m, d2=0.06m, d3=0.05mMa = 900 Kg, Mb = 700 Kgka = 0.85m, kb = 0.55m |
A. | 0.855m |
B. | 0.795m |
C. | 0.695m |
D. | 0.595m |
Answer» B. 0.795m | |
6. |
From the following data, calculate natural frequency of free torsional vibrations in Hz.l1=0.6m, l2=0.5m, l3=0.4md1=0.095m, d2=0.06m, d3=0.05mMa = 900 Kg, Mb = 700 Kgka = 0.85m, kb = 0.55mC = 80 GN/m2 |
A. | 3.37 |
B. | 7.95 |
C. | 6.95 |
D. | 5.95 |
Answer» B. 7.95 | |
7. |
In a system with different shaft parameters, the longest shaft is taken for calculations. |
A. | True |
B. | False |
Answer» C. | |
8. |
From the following data, calculate the equivalent length of shaft in m.l1=0.6m, l2=0.5m, l3=0.4md1=0.095m, d2=0.06m, d3=0.05m |
A. | 8.95 |
B. | 7.95 |
C. | 6.95 |
D. | 5.95 |
Answer» B. 7.95 | |