Question Solved1 Answer rad sec = A second order measurement system has a natural frequency of an = 125 a sensitivity of K = 0.05 [mm], and a damping ratio of 3 = 0.12. If the system is subjected to a step input [F(t) = AU(t)] of A = 75 [mm], determine the 90% rise time and the E5% settling time for a damping ratio of} = 0.15. Plot the response curve and indicate the corresponding rise and settling times.

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Transcribed Image Text: rad sec = A second order measurement system has a natural frequency of an = 125 a sensitivity of K = 0.05 [mm], and a damping ratio of 3 = 0.12. If the system is subjected to a step input [F(t) = AU(t)] of A = 75 [mm], determine the 90% rise time and the E5% settling time for a damping ratio of} = 0.15. Plot the response curve and indicate the corresponding rise and settling times.
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Transcribed Image Text: rad sec = A second order measurement system has a natural frequency of an = 125 a sensitivity of K = 0.05 [mm], and a damping ratio of 3 = 0.12. If the system is subjected to a step input [F(t) = AU(t)] of A = 75 [mm], determine the 90% rise time and the E5% settling time for a damping ratio of} = 0.15. Plot the response curve and indicate the corresponding rise and settling times.
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GIVEN:-Natural Frequency, =omega_(n)=125rod//sec{:[" Sensitivity, "K=0.05v//mm],[F(t)=A uu(t)],[A=75mm],[" domping Ratio "(xi_(i))=0.5]:}Solution:-T_(r)=(pi*theta)/(omega d)where{:[theta=cos^(-1)],[theta=cos^(-1)(0.15)],[theta=81.37]:}Converting into radion{:[theta=81.31 xx(pi)/(180)],[theta=0.452 pi]:}omega_(c)=omega_(n)sqrt(1-(xi)^(2))=125sqrt(1-(0.15)^(2))=123.58rad//secT_(r)=(pi-0.452 pi)/(123.58)T_(x)=(0.548%)/(123.58)For,xi=0.12T_(r)=(pi-theta)/(omega_(d) ... See the full answer