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Question-209430




Question Number 209430 by Tawa11 last updated on 09/Jul/24
Answered by mr W last updated on 10/Jul/24
1)  ω_0 =(√(k/m))  ζ=(c/(2(√(mk))))  ω_r =ω_0 (√(1−2ζ^2 ))=(√((k/m)(1−(c^2 /(2mk)))))       =(√(((12)/3)×(1−(8^2 /(2×3×12)))))=0.667 rad/s    2)  z_m =(√(c^2 +((k/ω)−mω)^2 ))       =(√(8^2 +(((12)/(15))−3×15)^2 ))=44.92 Ω
$$\left.\mathrm{1}\right) \\ $$$$\omega_{\mathrm{0}} =\sqrt{\frac{{k}}{{m}}} \\ $$$$\zeta=\frac{{c}}{\mathrm{2}\sqrt{{mk}}} \\ $$$$\omega_{{r}} =\omega_{\mathrm{0}} \sqrt{\mathrm{1}−\mathrm{2}\zeta^{\mathrm{2}} }=\sqrt{\frac{{k}}{{m}}\left(\mathrm{1}−\frac{{c}^{\mathrm{2}} }{\mathrm{2}{mk}}\right)} \\ $$$$\:\:\:\:\:=\sqrt{\frac{\mathrm{12}}{\mathrm{3}}×\left(\mathrm{1}−\frac{\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\mathrm{12}}\right)}=\mathrm{0}.\mathrm{667}\:{rad}/{s} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$${z}_{{m}} =\sqrt{{c}^{\mathrm{2}} +\left(\frac{{k}}{\omega}−{m}\omega\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\sqrt{\mathrm{8}^{\mathrm{2}} +\left(\frac{\mathrm{12}}{\mathrm{15}}−\mathrm{3}×\mathrm{15}\right)^{\mathrm{2}} }=\mathrm{44}.\mathrm{92}\:\Omega \\ $$
Commented by Tawa11 last updated on 10/Jul/24
Thanks for your time sir.  I really appreciate.
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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