SolSec3_1 - 3- 1Chapter Three Solutions Problem and...

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Unformatted text preview: 3- 1Chapter Three Solutions Problem and Solutions for Section 3.1 (3.1 through 3.14) 3.1Calculate the solution to !!x+2!x+2x=!t"#( )x( )=1!x( )=and plot the response. Solution: Given: !!x+2!x+2x=!t"#( )x( )=1,!x( )=!n=km=1.414 rad/s, "=c2km=0.7071, !d=!n1#"2=1 rad/sTotal Solution: x t( )=xht( )+xpt( )Homogeneous: From Equation (1.36) xht( )=Ae!"#ntsin#dt+$( )A=v+"#nx( )2+x#d( )2#d2, $=tan!1x#dv+"#nx%&’()*=.785 rad+xht( )=1.414e!tsint+.785( )Particular: From Equation. (3.9) xpt( )=1m!de"#!nt"$()sin!dt"$( )=11( )1( )e"t"%()sint"%( )But, sin"t( )="sintSo, xpt( )="e"t"%()sint&x t( )=1.414e!tsint+0.785( )<t<"x t( )=1.414e!tsint+0.785( )!e!(t!")sintt>"This is plotted below using the Heaviside function. 3- 23- 33.2Calculate the solution to !!x+2!x+3x=sint+!t"#( )x( )=!x( )=1and plot the response. Solution: Given:!!x+2!x+3x=sint+!t"#( ),x( )=0,!x( )=!n=km=1.732 rad/s, "=c2km=0.5774, !d=!n1#"2=1.414 rad/sTotal Solution: x t( )=xh+xp1<t<!x t( )=xh+xp1+xp2t>!Homogeneous: Eq. (1.36) xht( )=Ae!"#ntsin#dt+$( )=Ae!tsin 1.414t+$( )Particular: #1 (Chapter 2) xp1(t)=Xsin!t"#( ), where !=1 rad/s . Note that f=Fm=1$X=f!n2"!2( )2+2%!n!( )2=0.3536, and #=tan"12%!n!!n2"!2&’(()*++=0.785 rad$xp1t( )=0.3536sint"0.7854( )Particular: #2 Equation 3.9 xp2t( )=1m!de"#!nt"$()sin!dt"%( )=11( )1.414( )e"t"$()sin1.414t"$( )&xp2t( )=0.7071e"t"$()sin1.414t"$( )The total solution for 0< t<πbecomes: x t( )=Ae!tsin 1.414t+"( )+0.3536sint!0.7854( )!x t( )=!Ae!tsin(1.414t+")+1.414Ae!tcos 1.414t+"( )+0.3536cost!0.7854( )x( )==Asin"!0.25#A=0.25sin"!x( )=1=!Asin"+1.414Acos"+0.25#0.75=0.25!1.414 0.25( )1tan"#"=0.34 and A=0.75Thus for the first time interval, the response is x t( )=0.75e!tsin 1.414t+0.34( )+0.3536sint!0.7854( )<t<"Next consider the application of the impulse at t= π: 3- 4x t( )=xh+xp1+xp2x t( )=!0.433e!tsin 1.414t+0.6155( )+0.3536sint!0.7854( )!0.7071e!t!"()sin 1.414t!"( )t>"The response is plotted in the following (from Mathcad): 3.3Calculate the impulse response function for a critically damped system. Solution: The change in the velocity from an impulse isv=ˆFm, while x= 0. So for a critically damped system, we have from Eqs. 1.45 and 1.46 with x= 0: x(t)=vte!"nt#x(t)=ˆFmte!"nt3- 53.4Calculate the impulse response of an overdamped system. Solution: The change in velocity for an impulsev=ˆFm, while x= 0. So, for an overdamped system, we have from Eqs. 1.41, 1.42 and 1.43: x t( )=e!"#nt!v2#n"2!1e!#(n"2!1)t+v2#n"2!1e!#(n"2!1)t$%&&’())x t( )=ˆF2m#n"2!1e!"#nte!#(n"2!1)t!e!#(n"2!1)t$%&’()3.5 Derive equation (3.6) from equations (1.36) and (1.38). Solution: Equation 1.36: x(t) = Ae!"#ntsin#dt+$( )Equation 1.38: A=v+!"nx( )2+x"d( )2"d2,#=tan$1x"dv+!"nx%&’()*Since x= 0 and v= ˆFm,Equation 1.38 becomes A=v!...
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This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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SolSec3_1 - 3- 1Chapter Three Solutions Problem and...

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