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**Example text**

8. Free Vibration with Viscous Damping. Consider again the vibration of the system shown in Fig. 1 and assume that the vibrating body encounters in its motion a resistance proportional to the velocity. In such case, instead of equation of motion (1), we obtain W Wz = w- (W + fee) - ex. (a) ff The last term on the right side of this equation represents the force, proportional to velocity x. The minus damping sign shows that the force is acting in the direction opposite to the velocity. The coefficient c is a constant depending on the kind of the damping device and numerically is equal to the magnitude of the damping force when the velocity is equal to unity.

R-axis of all these forces P sin is zero, hence W cot kxi cxi x\ = 0, (k) Q the same equation as equation (32). This equation holds for any cot, hence the geometrical sum of the four vectors, shown in Fig. 28, is zero and the sum of their projections on any axis must be zero. Making projections on the directions Om and On we obtain which is value of the angle W Au + P cos a g 2 cAco + kA = 0, Psin a = 0. VIBRATION PROBLEMS IN ENGINEERING 44 From (33) these equations A and a. can be readily calculated and the formulae (34) for the amplitude of forced vibration and for the phase differ- and ence can be obtained.

Is acting on the load. , P = 2 Ibs. 56. 2 in. and the amplitude of forced From Solution. w = = 27T-5 vibration 2. is 31. 56 Determine the instant = t eq. 1 sec. if p (2), 'V ~g/8 8i . 128 in. W of the previous problem at the total displacement of the load at the initial moment (t = 0) the load at rest in equilibrium is position. 14 inch. 3. Determine the amplitude of forced torsional vibration of a shaft in Fig. 01 of a radian. Solution. Equation of motion in this case is (see Art. 2) M 1 1 , where

### ASHRAE 4146..Rates of Evaporation from Swimming Pools in Active Use

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