ASE 365 - HW 5 Solutions

ASE 365 - HW 5 Solutions - Solutions: Homework Set 5 1....

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Unformatted text preview: Solutions: Homework Set 5 1. Problem 4.5. Since F ( t ) = F T [ tu ( t )- ( t- T ) u ( t- T )], the response is x ( t ) = F T [ r ( t )- r ( t- T )] = F kT braceleftbiggbracketleftbigg t- 2 ζ ω n + e − ζω n t parenleftbigg 2 ζ ω n cos ω d t + 2 ζ 2- 1 ω d sin ω d t parenrightbiggbracketrightbigg u ( t )- bracketleftbigg t- T- 2 ζ ω n + e − ζω n ( t − T ) parenleftbigg 2 ζ ω n cos ω d ( t- T ) + 2 ζ 2- 1 ω d sin ω d ( t- T ) parenrightbiggbracketrightbigg · u ( t- T ) bracerightbigg For the undamped case, this reduces to x ( t ) = F kT braceleftbiggbracketleftbigg t- 1 ω n sin ω n t bracketrightbigg u ( t )- bracketleftbigg t- T- 1 ω n sin ω n ( t- T ) bracketrightbigg u ( t- T ) bracerightbigg which agrees with the results from the lecture. 2. Problem 4.7. The response to the trapezoidal pulse is x ( t ) = 2 F T bracketleftbigg r ( t )- r ( t- T 2 )- r ( t- 3 T 2 ) + r ( t- 2 T ) bracketrightbigg where r ( t ) is the ramp response found for problem 4.4, given by r ( t ) = 1 k bracketleftbigg t- 2 ζ ω n + e − ζω n t parenleftbigg 2 ζ ω n cos ω d t + 2 ζ 2- 1 ω d sin ω d t parenrightbiggbracketrightbigg u ( t ) 3. Problem 4.9. EOM: ¨ x + 2 ζω n ˙ x + ω 2 n x = F ( t ) m . Use second form from: x ( t ) = integraldisplay t F ( τ ) g ( t- τ ) dτ = integraldisplay t F ( t- τ ) g ( τ ) dτ where F ( t ) = F e − αt u ( t ) , g ( t ) = 1 mω d e − ζω n t sin ω d tu ( t ). Response: x ( t ) = F mω d integraldisplay t e − α ( t − τ ) u ( t- τ ) e − ζω n τ sin ω d τu ( τ ) dτ = F e − αt mω d integraldisplay t e ( α − ζω n ) τ sin ω d τ dτ = F m [ ω 2 d + ( α- ζω n ) 2 ] bracketleftbigg e − αt + parenleftbigg α- ζω n ω d sin ω d t- cos ω d t parenrightbigg e − ζω n t bracketrightbigg u ( t ) 1 With the provided values, this becomes x ( t ) = (0 . 041771 m) e ( − 1 sec- 1 ) t (1- cos(19 . 975 sec − 1 ) t ) u ( t ) . (In this result, units are expressed explicitly and carefully with the assumption that the textbook should have given α the units sec − 1 (or rad/sec).) 4. Problem 4.13. Using these two integrals found in the lecture: integraldisplay sin ω n ( t- τ ) dτ = 1 ω n cos ω n ( t- τ ) integraldisplay τ sin ω n ( t- τ ) dτ = 1 ω 2 n sin ω n ( t- τ ) + τ ω n cos ω n ( t- τ ) the results for the various time intervals can be found readily. For 0 < t < T/ 2: x 1 ( t ) = integraldisplay t 2 F T τ 1 mω n sin ω n ( t- τ ) dτ = 2 F mω n T bracketleftbigg 1 ω 2 n sin ω n ( t- τ ) + τ ω n cos ω n ( t- τ ) bracketrightbigg t = 2 F kT bracketleftbigg- 1 ω n sin ω n t + t bracketrightbigg For T/ 2 < t < 3 T/ 2: x 2 ( t ) = integraldisplay T/ 2 2 F T τ 1 mω n sin ω n ( t- τ ) dτ...
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This note was uploaded on 09/18/2011 for the course ASE 365 taught by Professor Staff during the Spring '10 term at University of Texas.

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ASE 365 - HW 5 Solutions - Solutions: Homework Set 5 1....

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