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1. A uniform bar of length L and weight W is suspended symmetrically by two strings as shown in Fig.1 Set up the differential equation of motion for small angular oscillations of the bar about the vertical axis z, and determine its period.

185_Figure 1.jpg

2. Set up the differential equation of motion for the system shown in Fig.2. Determine the expression for (a) the critical damping coefficient. And (b) the natural frequency of damped oscillation.

1627_Figure 2.jpg

3. A plate of area A and weight W is attached to the end of a spring and is allowed to oscillate in a viscous fluid as shown in Fig.3. If T1 is the natural period of undamped oscillation(that is, with the system oscillatingin air), and T2 the damped period with the plate immersed in the fluid. Determine the damping coefficient  μ,where the damping force on the plate is Fd = 2μAv, 2A is the total surface area of the plate, and v is its velocity.

1068_Figure 3.jpg

4. An undamped spring-mass system, m, k, is given a force excitation F(t) as shown in Fig. Determine its displacement responses of the system.

1328_Figure 4.jpg

5. Consider the following two degree of freedom vibration system.

529_Figure 5.jpg

The x1(t), x2(t) and x3(t) are the coordinates that measure the response of mass m1, m2 and m3.

1) First, determine the natural frequencies and mode shapes of this vibratory system. Obtain the natural frequencies and draw the

corresponding mode shapes, and explain why these two mode shapes have distinctly different characteristics.
2) If there is an external excitation on the 2st mass m2, which can be expressed as Feiωt. What will be the responses of the three masses ?

6. Fig.6 shows a string of length l and mass per unit length ρ is under tension T, and its upper end fixed at x = 0 and the lower end at x = l on a mass m which is supported on a spring of stiffness k and can slide in a slot in the transverse direction. Determine the equation for natural frequencies. Boundary conditions in this case are given by w(x,t)|x=0 = 0 and

-{T(∂w(x, t)/∂x) + kw(x, t)}|x=l = m(∂2w(x, t)/∂t2)|x=1

2356_Figure 6.jpg

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