1
$\begingroup$

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.

Improve this question
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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!

Improve this answer
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

$Mx''+cx'+kx=0$

$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.

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

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.