NATURAL frequency \(\omega_n = \SQRT {\frac{{{k_{eq}}}}{m}} \)\(f= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{eq}}}}{m}} \)Calculation:Given:k1 = k2 = 20 kN/m, m = 100 kgThis combination is a case of PARALLEL combination.keq = k1 + k2 = 40 kN/m = 40000 N/m\(f= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{eq}}}}{m}} \)\(= \frac{1}{{2\pi }}\sqrt {\frac{{40 \times {{10}^3}}}{{100}}} = \frac{{20}}{{2\pi }} = \frac{{10}}{\pi }\;Hz\)

"> NATURAL frequency \(\omega_n = \SQRT {\frac{{{k_{eq}}}}{m}} \)\(f= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{eq}}}}{m}} \)Calculation:Given:k1 = k2 = 20 kN/m, m = 100 kgThis combination is a case of PARALLEL combination.keq = k1 + k2 = 40 kN/m = 40000 N/m\(f= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{eq}}}}{m}} \)\(= \frac{1}{{2\pi }}\sqrt {\frac{{40 \times {{10}^3}}}{{100}}} = \frac{{20}}{{2\pi }} = \frac{{10}}{\pi }\;Hz\)

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A mass of 100 kg is held between two springs as shown in figure. The natural frequency of vibration of the system in cycles/second is

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

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Concept:∴ NATURAL frequency \(\omega_n = \SQRT {\frac{{{k_{eq}}}}{m}} \)\(f= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{eq}}}}{m}} \)Calculation:Given:k1 = k2 = 20 kN/m, m = 100 kgThis combination is a case of PARALLEL combination.keq = k1 + k2 = 40 kN/m = 40000 N/m\(f= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{eq}}}}{m}} \)\(= \frac{1}{{2\pi }}\sqrt {\frac{{40 \times {{10}^3}}}{{100}}} = \frac{{20}}{{2\pi }} = \frac{{10}}{\pi }\;Hz\)

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Posted on 31 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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