MECHANICAL VIBRATIONS PAPER

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MECHANICAL VIBRATIONS ME 408 May 2k6

Max Marks 60

Note: Question No. 1 is compulsory. Attempt any four questions from section B and two from section C.

Section A

I(a) What is meant by Natural Vibration?

(b) Define Resonance

(c) Mention important types of Free Vibrations.

(d) What is meant by Viscous damping?

(e) Define Vibration Isolation.

(f) Mention the uses of Lanchester damper

(g) What is an accelerometer and what is its use?

(h) Define influence coefficients.

(i) What is whirling speed?

(j) What is continuous system?

Section B

II Derive suitable expression for longitudinal vibrations for a rectangular uniform cross sectioned bar of length ‘l’ fixed at one end and free at the other end.

III Explain any two methods of vibration analysis.

IV Determine the effect of the mass of the spring on the natural frequency of the system shown in fig. 1 below

V A body of 5 kg is supported on a spring of stiffness 200 N/m has a dashpot connected to it which produces a resistance of 0.002 N at a velocity of 1 cm/sec. In what ratio will the amplitude of vibration be reduced after 5 cycles?

VI Solve the problem shown in figure below:m = 10 kg m2= 15 kg and k = 320 N/m

Section C

VII An electric motor is supported on a spring and a dashpot . the spring has the stiffness of 64 N/m and the dashpot offers resistance of 500 at 4 m/sec. The unbalanced mass of 0.5 kg rotates at 5 cm radius and the total mass of vibration system is 20 kg. The motor runs at 400 rpm. Determine a) damping factor b) amplitude vibration and phase angle c) resonant speed d) resonant amplitude and e) forces exerted by the spring and the motor..

VIII Two equal masses of weight 400 kg each and radius of gyration 40 cm are keyed to the opposite ends of a shaft 60 cm long. The shaft is 7.5 cm diameter for the first 25 cm of its length, 12.5 cm diameter for the next 10 cm and 8.5 cm for the remaining of its length. Find the frequency of free torsional vibration of the system and position of nodes. Assume G = 0.84 x 1011 N/m2.

IX Use Stodola’s method to find natural frequency of the system shown in Figure 3:

Take: E= 1.96 x 10 11 N/m2.

I = 4 x 10 -7 m2

m1= 160 kg, m2= 50 kg

0.3m