Sunday, July 19, 2020

2 Degree of Freedom System Vibration




Transfer Function at m1 (assuming no Force at m2)

$${\frac {v_1} {F_1}}=\frac {({{m_2s^3} +  {(k_2+k)s}})} {m_2m_1s^4+(m_2(k+k_1)+m_1(k+k_2))s^2 +    (k_2k_1+kk_1+k_2k)}   $$

Poles

$$ s = {\pm\sqrt{\frac {\left[{\pm\sqrt{\left[\frac {(k m_2 + k m_1 + k_2 m_1 + k_1 m_2)^2 - 4 m_2 m_1 (k k_2 + k k_1 + k_2 k_1)} {(m_2 m_1)}\right]} - \frac {k} {m_2} - \frac {k} {m_1} - \frac {k_2} {m_2} - \frac {k_1} {m_1}} \right]} {2}}} $$

Transfer Function at m2 (assuming no Force at m1)


$${\frac {v_2} {F_2}}=\frac {({{m_1s^3} +  {(k_1+k)s}})} {m_1m_2s^4+(m_1(k+k_2)+m_2(k+k_1))s^2 +    (k_1k_2+kk_2+k_1k)}   $$

Poles

$$ s = {\pm\sqrt{\frac {\left[{\pm\sqrt{\left[\frac {(k m_1 + k m_2 + k_1 m_2 + k_2 m_1)^2 - 4 m_1 m_2 (k k_1 + k k_2 + k_1 k_2)} {(m_1 m_2)}\right]} - \frac {k} {m_1} - \frac {k} {m_2} - \frac {k_1} {m_1} - \frac {k_2} {m_2}} \right]} {2}}} $$


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