This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Î¶ Ï‰ n + e âˆ’ Î¶Ï‰ n t parenleftbigg 2 Î¶ Ï‰ n cos Ï‰ d t + 2 Î¶ 2 1 Ï‰ d sin Ï‰ d t parenrightbiggbracketrightbigg u ( t ) bracketleftbigg t T 2 Î¶ Ï‰ n + e âˆ’ Î¶Ï‰ n ( t âˆ’ T ) parenleftbigg 2 Î¶ Ï‰ n cos Ï‰ d ( t T ) + 2 Î¶ 2 1 Ï‰ d sin Ï‰ d ( t T ) parenrightbiggbracketrightbigg Â· u ( t T ) bracerightbigg For the undamped case, this reduces to x ( t ) = F kT braceleftbiggbracketleftbigg t 1 Ï‰ n sin Ï‰ n t bracketrightbigg u ( t ) bracketleftbigg t T 1 Ï‰ n sin Ï‰ 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 Î¶ Ï‰ n + e âˆ’ Î¶Ï‰ n t parenleftbigg 2 Î¶ Ï‰ n cos Ï‰ d t + 2 Î¶ 2 1 Ï‰ d sin Ï‰ d t parenrightbiggbracketrightbigg u ( t ) 3. Problem 4.9. EOM: Â¨ x + 2 Î¶Ï‰ n Ë™ x + Ï‰ 2 n x = F ( t ) m . Use second form from: x ( t ) = integraldisplay t F ( Ï„ ) g ( t Ï„ ) dÏ„ = integraldisplay t F ( t Ï„ ) g ( Ï„ ) dÏ„ where F ( t ) = F e âˆ’ Î±t u ( t ) , g ( t ) = 1 mÏ‰ d e âˆ’ Î¶Ï‰ n t sin Ï‰ d tu ( t ). Response: x ( t ) = F mÏ‰ d integraldisplay t e âˆ’ Î± ( t âˆ’ Ï„ ) u ( t Ï„ ) e âˆ’ Î¶Ï‰ n Ï„ sin Ï‰ d Ï„u ( Ï„ ) dÏ„ = F e âˆ’ Î±t mÏ‰ d integraldisplay t e ( Î± âˆ’ Î¶Ï‰ n ) Ï„ sin Ï‰ d Ï„ dÏ„ = F m [ Ï‰ 2 d + ( Î± Î¶Ï‰ n ) 2 ] bracketleftbigg e âˆ’ Î±t + parenleftbigg Î± Î¶Ï‰ n Ï‰ d sin Ï‰ d t cos Ï‰ d t parenrightbigg e âˆ’ Î¶Ï‰ n t bracketrightbigg u ( t ) 1 With the provided values, this becomes x ( t ) = (0 . 041771 m) e ( âˆ’ 1 sec 1 ) t (1 cos(19 . 975 sec âˆ’ 1 ) t ) u ( t ) . (In this result, units are expressed explicitly and carefully with the assumption that the textbook should have given Î± the units sec âˆ’ 1 (or rad/sec).) 4. Problem 4.13. Using these two integrals found in the lecture: integraldisplay sin Ï‰ n ( t Ï„ ) dÏ„ = 1 Ï‰ n cos Ï‰ n ( t Ï„ ) integraldisplay Ï„ sin Ï‰ n ( t Ï„ ) dÏ„ = 1 Ï‰ 2 n sin Ï‰ n ( t Ï„ ) + Ï„ Ï‰ n cos Ï‰ n ( t Ï„ ) the results for the various time intervals can be found readily. For 0 < t < T/ 2: x 1 ( t ) = integraldisplay t 2 F T Ï„ 1 mÏ‰ n sin Ï‰ n ( t Ï„ ) dÏ„ = 2 F mÏ‰ n T bracketleftbigg 1 Ï‰ 2 n sin Ï‰ n ( t Ï„ ) + Ï„ Ï‰ n cos Ï‰ n ( t Ï„ ) bracketrightbigg t = 2 F kT bracketleftbigg 1 Ï‰ n sin Ï‰ n t + t bracketrightbigg For T/ 2 < t < 3 T/ 2: x 2 ( t ) = integraldisplay T/ 2 2 F T Ï„ 1 mÏ‰ n sin Ï‰ n ( t Ï„ ) dÏ„...
View
Full Document
 Spring '10
 Staff
 Sin, â„¦n, sin Ï‰n, mÏ‰n Ï‰n

Click to edit the document details