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so for the Single Degree of freedom system, I obtained the equation in blue, where I think the equation below in black is missing the ‘m*a’ from F=ma?

So I see that there is one ‘m*a’, but is that from the inertia or the F=ma?

Am I missing something here please? enter image description here

enter image description here

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  • $\begingroup$ Blue equation has $m\ddot{x}$ on both sides of the $=$ sign. They will cancel out. Blue equation is probably wrong. $\endgroup$
    – AJN
    Apr 11, 2022 at 12:33
  • $\begingroup$ Hi, is that ‘m*a’ from the black equation please from the inertia of F=ma? $\endgroup$
    – Batsova
    Apr 11, 2022 at 18:08
  • $\begingroup$ "I obtained the equation in blue...". Please show the steps by which you obtained the equation. Please use edit to add details directly into the question. $\endgroup$
    – AJN
    Apr 12, 2022 at 12:14

1 Answer 1

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The black equation is a rearranged form because $m\cdot a = m\cdot \ddot{x}$ (i.e. the acceleration is equation to the second derivative of position with respect to time) .

So (because positive is downwards) : $$\sum_F = m\cdot a \rightarrow$$ $$-k\cdot x -c\cdot \dot{x} + F(t)= m\cdot \ddot{x}\rightarrow$$

Then rearranging and putting all $x$ related variables to the right side:

$$m\cdot \ddot{x}+ c\cdot \dot{x} +k\cdot x = F(t) $$

The $F(t)$ is the external excitation force (the force that is applied and leads to the forced vibration).

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  • $\begingroup$ Hi, so was I wrong? I thought that ‘m*a’ in the equation was from the mass inertia? $\endgroup$
    – Batsova
    Apr 11, 2022 at 12:44
  • $\begingroup$ When representing $m\ddot{x}$ as a fictitious force you treat the system as if it is in static equilibrium, i.e. $\sum F=0$ $\endgroup$
    – drC1Ron
    Jun 4, 2022 at 20:18
  • $\begingroup$ @RonnyLandsverk since $\ddot{x}= a$ I would argue that I am using $\sum F = m\cdot a$ $\endgroup$
    – NMech
    Jun 5, 2022 at 6:22
  • $\begingroup$ @NMech Google D'Alembert's principle. Your free-body-diagram has $m\ddot{x}$ on it. So you cannot use $\sum F=ma$. You need to use $\sum F=0$ $\endgroup$
    – drC1Ron
    Jun 5, 2022 at 7:38
  • $\begingroup$ @NMech Btw, my first comment was intented for the OP and not specifically for your answer. $\endgroup$
    – drC1Ron
    Jun 5, 2022 at 7:48

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