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So we have a mass spring system, which is supposed to be modelling a bridge, and the equation that the displacement of the bridge $x$ satisfies is given by $$M\ddot{x}+c\dot{x}+kx=0$$ where $M=4\times 10^5\text{kg}, c=5\times 10^4\text{kg/s}, k=10^7\text{kg} $, where $m$ is the mass, $k$ is the 'spring constant'and $c$ is the friction or level of damping or something, I'm not sure - this equation is pretty standard for this type of problem so you probably know what it means more than me anyway.

Now there is a part of the question where another factor is introduced, so that $x$ now satisfies the equation $$M\ddot{x}+c\dot{x}+kx=300\dot{N}.$$

Now my question is, is the damping level a constant value, so is it still $c$ or has it changed due to the new factor affecting $\dot{x}$ so would the damping now be $c-300N$? or does it just stay at $c$?

I think the main problem here is that I don't really know what is damping (what actually is damping )in general, so if anyone could just answer the question I asked above or say something about damping in general in these types of problems that would explain my confusion.

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  • $\begingroup$ What is $N$ or $\dot N$ on the RHS? Damping essentially means that the amplitude of oscillations gets smaller and smaller as time evolves. Imagine the ODE without the term $c\dot x$, then you will have a harmonic oscillator where the amplitude doesn't decrease as time evolves; once you include the term $c\dot x$, the solution is multiplied by an exponential decaying function (see below), which is exactly what I just explained about damping. $\endgroup$ – Chee Han Nov 9 '16 at 8:24
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$Mx''+cx'+kx=0$

$x = Ae^{-\frac c{2M}t}\sin\big(\phi + t\sqrt {\frac {k}{M} -(\frac{c}{2M})^2} \big) $

I hope I have that right, I eyeballed it.

Now we change the equation. $Mx''+cx'+kx=300$

What happens to $x?$ $x$ adjusts by a constant.

$x = Ae^{-\frac c2t}\sin(\phi + t\sqrt {Mk -\frac {c^2}4} ) + K$

$x'$ and $x''$ are unchanged.

$kK = 300\\ K = \frac {300}{k}$

$x = Ae^{-\frac c2t}\sin(\phi + t\sqrt {Mk -\frac {c^2}4} ) + \frac {300}{k}$

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Damping in this model is a force that resists motion and that is proportional to the instantaneous velocity,

$$ F_{\rm damping} = -c \frac{dx}{dt} $$

This is a very typical force when you're immersed in a fluid. In this case $c$ is a constant, the larger $c$ the stronger is the force opposing the motion. So for instance between oil and water, oil has a larger $c$.

Now, there is also a restoring force in your model that is taken proportional to the deformation of the bridge

$$ F_{\rm restoring} = -k x $$

And finally an external force (e.g. wind) $F_{\rm external}$. Applying Newton's second law you find

\begin{eqnarray} Ma = \sum_i F_i &=& F_{\rm restoring} + F_{\rm damping} + F_{\rm external} \\ M \ddot{x} + a\dot{x} + kx &=& F_{\rm external} \end{eqnarray}

$c$ is still constant, regardless of the fact that the bridge is being externally forced to move!

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  • $\begingroup$ A later part of the question says that additional damping is added, does this mean that the value of $c$ increases since the level of damping has been increased artificially? $\endgroup$ – Anon Nov 9 '16 at 1:38
  • $\begingroup$ Correct, $c$ increases $\Rightarrow$ damping increases $\endgroup$ – caverac Nov 9 '16 at 1:42
  • $\begingroup$ That makes sense, so in the case of a bridge, $c$ is set to a level of whatever the designers of the bridge wanted, so they could just increase the value of $c$ if they wanted to by adding some structure to the bridge (I'm not engineer)? $\endgroup$ – Anon Nov 9 '16 at 2:16
  • $\begingroup$ This is a very simple model, but indeed, the geometry of the bridge impacts the effective value of $c$. The materials used to build it usually go into $k$ $\endgroup$ – caverac Nov 9 '16 at 2:19

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