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Changing the relative equilibrium position for a 2-mass spring-mass system

At equilibrium $\ddot{r}_2=0,\ \dot{r}_2=0$. So, from the second equation, $$r_2-r_1 = \frac{m_2 g \cos\theta}{K_{12}}$$ For positive values of $m, g, \cos\theta, K_{12}$, $r_2 > r_1$. i.e. mass #2 ...
AJN's user avatar
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1 vote
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What is the bandwidth for a 2nd order system?

This question is related to: Contradiction of bandwidth and damping We know that the transfer function for a second-order system is: $$ H(s) = \frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2} $...
rhody's user avatar
  • 121

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