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Question Number 209430 by Tawa11 last updated on 09/Jul/24

Answered by mr W last updated on 10/Jul/24

$$\left.\mathrm{1}\right) \\$$$$\omega_{\mathrm{0}} =\sqrt{\frac{{k}}{{m}}} \\$$$$\zeta=\frac{{c}}{\mathrm{2}\sqrt{{mk}}} \\$$$$\omega_{{r}} =\omega_{\mathrm{0}} \sqrt{\mathrm{1}−\mathrm{2}\zeta^{\mathrm{2}} }=\sqrt{\frac{{k}}{{m}}\left(\mathrm{1}−\frac{{c}^{\mathrm{2}} }{\mathrm{2}{mk}}\right)} \\$$$$\:\:\:\:\:=\sqrt{\frac{\mathrm{12}}{\mathrm{3}}×\left(\mathrm{1}−\frac{\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\mathrm{12}}\right)}=\mathrm{0}.\mathrm{667}\:{rad}/{s} \\$$$$\\$$$$\left.\mathrm{2}\right) \\$$$${z}_{{m}} =\sqrt{{c}^{\mathrm{2}} +\left(\frac{{k}}{\omega}−{m}\omega\right)^{\mathrm{2}} } \\$$$$\:\:\:\:\:=\sqrt{\mathrm{8}^{\mathrm{2}} +\left(\frac{\mathrm{12}}{\mathrm{15}}−\mathrm{3}×\mathrm{15}\right)^{\mathrm{2}} }=\mathrm{44}.\mathrm{92}\:\Omega \\$$

Commented by Tawa11 last updated on 10/Jul/24

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\$$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\$$