PARALLEL, so \({k_{EQ}} = {k_1} + {k_2} + {k_3} + {k_4} = 16 + 16 + 32 + 32 \RIGHTARROW {K_{eq}} = 96\frac{{MN}}{m}\)we know that at resonance \(\omega = {\omega _n} = \sqrt {\frac{k}{m}} \Rightarrow \frac{{2\pi N}}{{60}} = \sqrt {\frac{{{k_{eq}}}}{m}}\)\(\Rightarrow N = \frac{{60}}{{2\pi }}\times \sqrt {\frac{{96 \times {{10}^6}}}{{240}}} = 6039.5 \simeq 6040\ RPM\)

"> PARALLEL, so \({k_{EQ}} = {k_1} + {k_2} + {k_3} + {k_4} = 16 + 16 + 32 + 32 \RIGHTARROW {K_{eq}} = 96\frac{{MN}}{m}\)we know that at resonance \(\omega = {\omega _n} = \sqrt {\frac{k}{m}} \Rightarrow \frac{{2\pi N}}{{60}} = \sqrt {\frac{{{k_{eq}}}}{m}}\)\(\Rightarrow N = \frac{{60}}{{2\pi }}\times \sqrt {\frac{{96 \times {{10}^6}}}{{240}}} = 6039.5 \simeq 6040\ RPM\)

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An automotive engine weighing 240 kg is supported on four springs with linear characteristics. Each of the front two springs have a stiffness of 16 MN/m while the stiffness of each rear spring is 32 MN/m. The engine speed (in rpm), at which resonance is likely to occur, is

Mechanical Vibrations Undamped Free Vibration in Mechanical Vibrations . 7 months ago

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k1 = k2 = 16 MN/mk3 = k4 = 32 MN/m, m = 240 kgsprings are PARALLEL, so \({k_{EQ}} = {k_1} + {k_2} + {k_3} + {k_4} = 16 + 16 + 32 + 32 \RIGHTARROW {K_{eq}} = 96\frac{{MN}}{m}\)we know that at resonance \(\omega = {\omega _n} = \sqrt {\frac{k}{m}} \Rightarrow \frac{{2\pi N}}{{60}} = \sqrt {\frac{{{k_{eq}}}}{m}}\)\(\Rightarrow N = \frac{{60}}{{2\pi }}\times \sqrt {\frac{{96 \times {{10}^6}}}{{240}}} = 6039.5 \simeq 6040\ RPM\)

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Posted on 26 Oct 2024, this text provides information on Mechanical Vibrations related to Undamped Free Vibration in Mechanical Vibrations. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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