JNTU
JNTU B.Tech IV MECHANICAL VIBRATIONS AND STRUCTURAL DYNAMICS Previous Question Paper (July - 2021 - R16)
Answer any Five Questions. All Questions Carry Equal Marks.
Main
Derive the expressions for under damping, critical damping and over damping of free damped vibrations of a single degree freedom system.
Answer
For a single degree of freedom system with viscous damping, the equation of motion is given by:
Dividing by mass m, we get:
where ωn = √(k/m) is the natural frequency and ζ = c / (2mωn) is the damping ratio. The characteristic equation is s2 + 2ζωns + ωn2 = 0, with roots:
The nature of the motion depends on the value of ζ:
- Under Damping (ζ < 1): The roots are complex conjugates: s1,2 = -ζωn ± iωd, where ωd = ωn√(1 - ζ2) is the damped natural frequency. The displacement is: x(t) = e-ζωnt(A cos ωdt + B sin ωdt)This represents an oscillatory motion with decaying amplitude.
- Critical Damping (ζ = 1): The roots are real and equal: s1,2 = -ωn. The displacement is: x(t) = (A + Bt)e-ωntThis provides the fastest return to equilibrium without oscillation.
- Over Damping (ζ > 1): The roots are real and distinct: s1,2 = -ζωn ± ωn√(ζ2 - 1). The displacement is: x(t) = Aes1t + Bes2tThis results in a non-oscillatory, slow return to the equilibrium position.
Determine the equivalent spring constant of the system shown in Figure 1.
Answer
The equivalent spring constant depends on the configuration of the springs. For springs in parallel, the equivalent stiffness is the sum of individual stiffnesses:
Find the natural frequency of the pulley system shown in Figure 2 by neglecting the friction and the masses of the pulleys.
Answer
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Derive and solve the equation of motion with harmonic force system shown in Figure 3.
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Derive the equation of motion of a system subjected to base excitation.
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Find the solution of the following equation when the forcing function is a unit impulse at t = 0 and determine the initial and steady-state values of the response.
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Calculate the natural frequency of a shaft of diameter 10 cm and length 300 cm carrying two discs of diameters 125 cm and 200 cm respectively at its ends and weighing 480 N and 900 N respectively. Modulus of rigidity of the shaft may be taken as 2 × 105 N/mm2.
Answer
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Consider the two degree-of-freedom system of Figure 4. Determine the steady-state response of the system.
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A steel wire of 2 mm diameter is fixed between two points located 2 m apart. The tensile force in the wire is 250 N. Determine the fundamental frequency of vibration and the velocity of wave propagation in the wire.
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A uniform bar of cross-sectional area A, length l, and Young’s modulus E is connected at both ends by springs, dampers, and masses, as shown in Figure 5. State the boundary conditions.
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