rad/sec Example 2.13: Find out the equation of motion for the vibratory system shown in fig Sample copy

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Natural frequency, ω = rad/sec So, ω = rad/sec Example 2.13: Find out the equation of motion for the vibratory system shown in fig 2.26. Fig 2.

1 Natural frequency, ω = rad/sec So, ω = rad/sec Example 2.13: Find out the equation of motion for the vibratory system shown in fig Fig 2.26 Solution: The equation of motion can be written as (ka)aθ + (mb)bθ = F sin ωt. b ka θ + mb θ = F sin ωt. b Example 2.14: Derive the differential equation of motion for a spring controlled simple pendulum as shown in fig.2.36 The spring is in its up stretched position when the pendulum rod is vertical. Solution: let us say the system is displaced by an angle θ to the right. Iθ = mgl θ ka. aθ ml θ + (mgl + ka )θ = 0(I = ml ) θ + θ = 0 θ + + θ = 0 Fig 2.36 ω = + rad/sec

2 Example 2.15: A 5 kg mass attached to the lower end of spring, whose upper end is fixed, vibrates with a natural period of 0.45 sec. Determine the natural period when a 2.5 kg mass is attached to the midpoint of the same spring with the upper and lower ends fixed. Solution: Natural frequency can be Or Fig 2.36A ω = =. ω = k m = = rad/sec (13.95) = k = m(13.95) = 5(13.95) = 973 N/m When the spring is divided into two parts, its stiffness will be twice i.e. 2k, but now two parts of the spring are in parallel so k = 4k. Then ω = = = rad/sec. Natural period =. = sec

3 Example 2.16: A steel wire with E = N/m 2 is of 2 mm diameter and is 30 mm long. It is fixed at the upper end and carries a mass M kg at its lower end. Find M so that the frequency of longitudinal vibrations is 4 cycles/sec. Solution: f = 1 2π k m = 4 cycle sec = 4 Hz For cantilever Putting in (i) Putting in (ii) k = δ = static de lection = f = E = N/m I = d = l = 30 mm = = 3EI = = m 4 = 1 2π (30) M (i) (ii) M = Kg Example 2.16: Determine the natural frequency of the system shown in fig Solution: The total kinetic energy of the given system K. E = 1 2 mx Iθ Now x is related to θ as Hence x = 3rθ x = 3rθ So the kinetic energy = m 3rθ + Iθ = mr θ + Iθ = K. E P. E = 1 2 kx = 1 2 k(rθ) = 1 2 kr θ

4 Applying energy method d (K. e + P. E) = 0 dt d dt 9 2 mr θ Iθ kr θ = mr 2θ θ I 2θ θ kr 2θθ = 0 1) If the mass of a body increases 9 times a) Frequency decreases 3 times b) Frequency increases 3 times c) Frequency does not change d) It will be zero. 9 2 mr θ Iθ kr θ = mr + I θ + kr θ = 0 ω = kr 9mr + 1 Objective Type Questions 2) In Rayleigh's method for finding frequency of the system a) Max kinetic energy is equal to max potential energy b) Kinetic energy is more than potential energy c) Potential energy is zero d) Kinetic energy is zero 3) In energy method for finding frequency of the system a) The sum of kinetic and potential energy is zero b) The sum of kinetic and potential energy is constant c) It is assumed that the system is non-conservative d) Frequency cannot be determined by energy method 4) Bifilar suspension can be used to find the

5 a) Moment of inertia of bar b) M.I of the disc c) Mass of the disc d) Mass of the bar 5) Trifilar suspension is used to find a) Moment of inertia of bar b) M.I of disc c) Mass of the disc d) Mass of the bar 6) The main properties of a vibrating system are a) Mass and stiffness b) Mass, stiffness and damping c) Stiffness and damping d) Damping and stiffness 7) The natural frequency of a system is function of a) The stiffness of the system b) The mass of the system c) Both a and b d) None of the above 8) In the spring mass system if the mass of the system is doubled with spring stiffness halved, the natural frequency of longitudinal vibration a) Remained unchanged b) Is doubled c) Is halved d) Is quadrupled 9) If the spring mass system with m and spring stiffness k is taken to very high altitude, the natural frequency of longitudinal vibrations a) Increases b) decreases c) remains unchanged d) may increase or decrease depending upon the value of the mass

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Engineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati Introduction to vibration

Engineering Mechanics Prof. U. S. Dixit Department of Mechanical Engineering Indian Institute of Technology, Guwahati Introduction to vibration Module 15 Lecture 38 Vibration of Rigid Bodies Part-1 Today,