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# SolSec1_2to1_3 - Problems and Solutions for Section 1.2 and...

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Problems and Solutions for Section 1.2 and Section 1.3 (1.20 to 1.51) Problems and Solutions Section 1.2 (Numbers 1.20 through 1.30) 1.20* Plot the solution of a linear, spring and mass system with frequency ω n =2 rad/s, x 0 = 1 mm and v 0 = 2.34 mm/s, for at least two periods. Solution: From Window 1.18, the plot can be formed by computing: A = 1 ! n ! n 2 x 0 2 + v 0 2 = 1.54 mm, " = tan # 1 ( ! n x 0 v 0 ) = 40.52 ! x ( t ) = A sin( ! n t + " ) This can be plotted in any of the codes mentioned in the text. In Mathcad the program looks like. In this plot the units are in mm rather than meters.

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1.21* Compute the natural frequency and plot the solution of a spring-mass system with mass of 1 kg and stiffness of 4 N/m, and initial conditions of x 0 = 1 mm and v 0 = 0 mm/s, for at least two periods. Solution: Working entirely in Mathcad, and using the units of mm yields: Any of the other codes can be used as well.
1.22 To design a linear, spring-mass system it is often a matter of choosing a spring constant such that the resulting natural frequency has a specified value. Suppose that the mass of a system is 4 kg and the stiffness is 100 N/m. How much must the spring stiffness be changed in order to increase the natural frequency by 10%? Solution: Given m =4 kg and k = 100 N/m the natural frequency is ! n = 100 4 = 5 rad/s Increasing this value by 10% requires the new frequency to be 5 x 1.1 = 5.5 rad/s. Solving for k given m and ω n yields: 5.5 = k 4 ! k = (5.5) 2 (4) = 121 N/m Thus the stiffness k must be increased by about 20%.

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1.23 Referring to Figure 1.8, if the maximum peak velocity of a vibrating system is 200 mm/s at 4 Hz and the maximum allowable peak acceleration is 5000 mm/s 2 , what will the peak displacement be? mm/sec 200 = v x (mm) a = 5000 mm/sec 2 f = 4 Hz Solution: Given: v max = 200 mm/s @ 4 Hz a max = 5000 mm/s @ 4 Hz x max = A v max = A ω n a max = A ω n 2 ! x max = v max " n = v max 2 # f = 200 8 # = 7.95 mm At the center point, the peak displacement will be x = 7.95 mm
1.24 Show that lines of constant displacement and acceleration in Figure 1.8 have slopes of +1 and –1, respectively. If rms values instead of peak values are used, how does this affect the slope? Solution: Let x = x max sin ! n t ˙ x = x max ! n cos ! n t ˙ ˙ x = " x max ! n 2 sin ! n t Peak values: ! x max = x max ! n = 2 " fx max !! x max = x max ! n 2 = (2 " f ) 2 x max Location: f x x f x x ! ! 2 ln ln ln 2 ln ln ln max max max max " = + = ! ! ! ! Since x max is constant, the plot of ln max x ! versus ln 2 π f is a straight line of slope +1. If ln max x ! ! is constant, the plot of ln max x ! versus ln 2 π f is a straight line of slope –1. Calculate RMS values Let x t ( ) = A sin ! n t ˙ x t ( ) = A ! n cos ! n t ˙ ˙ x t ( ) = " A ! n 2 sin ! n t

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Mean Square Value: x 2 = T ! " lim 1 T x 2 0 T # ( t ) dt x 2 = T ! " lim 1 T A