I get stuck in an ODE exercise. I have only studied Physics in secondary school, so I have a hard time understanding the question:
Oscillations on a foot bridge can be modeled by the following ODE:
$$M x'' + k x' + h x = 300 N x'$$
where $M, k, h \in \mathbb{R}^+$ and $N \in \mathbb{N}$.
Physically, $M$ is the mass and $N$ is the number of pedestrians walking on the bridge.
Our goal is to express $N_0 = \min \{N_1 \in \mathbb{N} \mid \text{the bridge is no longer damped if $N > N_1$}\}$ as a function of $k, h, M$.
What is meant by $``\text{the bridge is no longer damped}''$? Thank you.
It means that the amplitude of the solution diverges at the increasing of time.
So, recast first the equation in canonical form $$ \eqalign{ & Mx'' + kx' + hx = 300Nx' \cr & Mx'' + \left( {k - 300N} \right)x' + hx = 0 \cr & x'' + \left( {{{k - 300N} \over M}} \right)x' + {h \over M}x = 0 \cr} $$ which is a 2nd order linear ODE.
Then you (presumably) know that the solution to the linear ODE $$ x'' + a\,x' + b\,x = 0 $$ is $$ \eqalign{ & x(t) = c_{\,1} \,e^{ - \,a/2\,t} \,e^{\left( {\sqrt {a^{\,2} - 4b} } \right)/2\;t} + c_{\,2} \,e^{ - \,a/2\,t} \,e^{ - \,\left( {\sqrt {a^{\,2} - 4b} } \right)/2\;t} = \cr & = e^{ - \,a/2\,t} \left( {c_{\,1} \,\,e^{\left( {\sqrt {a^{\,2} - 4b} } \right)/2\;t} + c_{\,2} \,e^{ - \,\left( {\sqrt {a^{\,2} - 4b} } \right)/2\;t} } \right) \cr} $$
The evolution in time of the amplitude of the signal is governed by $$ e^{ - \,a/2\,t} $$ irrespective of whether the square root provides real or imaginary values, because:
So the amplitude of the resulting signal will damp out in time if $0<a$ and $0<b$.
In our case $$ a = {{k - 300N} \over M} $$ and $h$ and $M$ and thus $b$ are positive, for physical considerations.
Finally we get that the system becomes unstable = no longer damped if $$ a \le 0\quad \Rightarrow \quad {{k - 300N} \over M}\le 0\quad \Rightarrow \quad {k \over {300}} \le N\quad \Rightarrow \quad \left\lceil {{k \over {300}}} \right\rceil \le N $$