SolSec1.2_1.3 - Problems and Solutions for Section 1.2 and...

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Problems and Solutions for Section 1.2 and Section 1.3 (1.18 to 1.46) Problems and Solutions Section 1.2 (Numbers 1.18 through 1.26) 1.18* 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 first changing the units to meters so that xv n 00 0 001 0 00234 2 == = .. m, m/s and rad/s ω . Then Ax v x v n n n =+ = = = 1 154 4052 2 0 2 0 21 0 0 ωφ . tan ( ) . mm, o xt A t n ( ) sin( ) 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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1.19* 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.
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1.20 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 ω 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: 55 4 5 5 4 121 2 .( . ) ( ) =⇒ = = k k N/m Thus the stiffness k must be increased by about 20%.
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1.21 Referring to Figure 1.7, 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 n 2 95 . 7 8 200 2 max max max = = = = π f v v x At the center point, the peak displacement will be x = 7.95 mm
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1.22 Show that lines of constant displacement and acceleration in Figure 1.7 have slopes of +1 and –1, respectively. If rms values instead of peak values are used, how does this affect the slope? Solution: Let xx t t t n nn = = =− max max max sin ˙ cos ˙˙ sin ω ωω 2 Peak values: ˙ () ˙ max max max max max max f x f x n n == ωπ 2 2 22 Location: f x x f x x π 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 xt A t A t A t n () = ()=− sin ˙ cos sin 2
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Mean Square Value: x T xtd t T T 22 0 1 = →∞ lim () x T At d t A T td t A T n T T n T 2 0 2 0 2 1 12 2 == = ∫∫ lim sin lim ( cos ) ωω x T d t A T t A T nn T T n n T n . lim cos lim ( cos ) 2 2 0 2 2 0 2 2 11 2 2 = =+ = ω x T d t A T t A T T T n n T n ..
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This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec1.2_1.3 - Problems and Solutions for Section 1.2 and...

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