# Forced vibration single degree of freedom system

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?

• Blue equation has $m\ddot{x}$ on both sides of the $=$ sign. They will cancel out. Blue equation is probably wrong.
– AJN
Apr 11, 2022 at 12:33
• Hi, is that ‘m*a’ from the black equation please from the inertia of F=ma? Apr 11, 2022 at 18:08
• "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.
– AJN
Apr 12, 2022 at 12:14

## 1 Answer

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).

• Hi, so was I wrong? I thought that ‘m*a’ in the equation was from the mass inertia? Apr 11, 2022 at 12:44
• When representing $m\ddot{x}$ as a fictitious force you treat the system as if it is in static equilibrium, i.e. $\sum F=0$ Jun 4, 2022 at 20:18
• @RonnyLandsverk since $\ddot{x}= a$ I would argue that I am using $\sum F = m\cdot a$
– NMech
Jun 5, 2022 at 6:22
• @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 $\sum F=0$ Jun 5, 2022 at 7:38
• @NMech Btw, my first comment was intented for the OP and not specifically for your answer. Jun 5, 2022 at 7:48