When first opened, the Millennium Bridge in London wobbled from side to side as people crossed; you can see this on video at www.arup.com/MillenniumBridge. Footfalls created small side-to-side movements of the bridge, which were then enhanced by the tendency of people to adjust their steps to compensate for wobbling. With more than a critical number of pedestrians (around 160) the bridge began to wobble violently. Without any pedestrians, the displacement x of a representative point on the bridge away from its normal position would satisfy $Mx'' + kx' + λx = 0$, where $M ≈ 4 × 10^5kg$, $k ≈ 5 × 10^4kg/s$, $λ ≈ 10^7kg/s$
Show that the level of damping here is only around 1% of the critical value
I did $k^2-4Mλ$ which is a large value nowhere close to 0. In the answer they work out the value of $k$ from $M$ and $λ$, knowing that $k^2-4Mλ=0$ and they show that they given value of k is around 1% to the worked out value of k. I dont understand how working out this answers the question as k is not the damping itself
In the absence of any damping, the bridge would oscillate with a frequency proportional to $\sqrt{\lambda/M}$. Note that the unit of $\lambda$ is $kg/s^2$ not $kg/s$. People don't like to walk on oscillating bridges, so the architect/constructor make sure that the oscillations are damped. Preferably, the damping factor should be at least equal to the critical damping. You calculate the critical damping by $k_c^2-4M\lambda=0$. The actual damping of the bridge in question was far less than the desired value. Additional dampers were added to the bridge later. See here, page 8.