Question

On June 10, 2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after only two days, it had to be closed to traffic for safety reasons. On the opening day, in fact, so many people were crossing it at the same time that unexpected sideways oscillations of the bridge were observed. Further investigations indicated that the oscillation was caused by lateral forces produced by the synchronization of steps taken by the pedestrians. Although the origin of this cadence synchronization was new to the engineers, its effect on the structure of the bridge was very well known. Thecombined forces exerted by the pedestrians as they were walking in synchronization had a frequency very close to the natural frequency of the bridge, and resonance occurred.

Consider an oscillating system of mass m and natural angular frequency ω_n. When the system is subjected to aperiodic external (driving) force, whose maximum value is F_max and angular frequency is ω_d, the amplitude of the driven oscillations is

\(A=\frac{F_{\max }}{\sqrt{\left(k-m \omega_{d}^{2}\right)^{2}+\left(b \omega_{d}\right)^{2}}}\),

where k is the force constant of the system and b is the damping constant.

We will use this simple model to study the oscillations of the Millennium Bridge.

Part A

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 N at a frequency of about 1 Hz. Given that the mass of the bridge is about 2000 kg per linear meter, how many people were walking along the 144-m-long central span of the bridge at one time, when an oscillation amplitude of 75 mm was observed in that section of the bridge? Take the damping constant to be such that the amplitude of the undriven oscillations would decay to 1/e of its original value in a time t=6T, where T is the period of the undriven, undamped system.

Answer ≈ 1810 people (to 3 significant figures).

Part B

What would the amplitude A of oscillation of the MillenniumBridge have been on the opening day if the damping effects had been three times more effective?

Answer 1

Concepts and reason

The concept required to solve the given problem is damped oscillation. First, calculate the damping constant by the equation of amplitude of wave at a given time $t$ . Next, calculate the number of people walking on the bridge by using the condition for which the amplitude will be maximum. Finally, calculate the maximum amplitude of the driven oscillation.

Fundamentals

The amplitude of undamped oscillation is given by,

$A = {A_0}{e^{ - \frac{{bt}}{{2m}}}}$

Here, $b$ is the damping constant, $m$ is the mass, $T$ is the time-period and ${A_0}$ is the amplitude at time $t = 0{\rm{ s}}$ .

When an oscillating system of mass $m$ and natural angular frequency ${\omega _n}$ is subjected to a periodic external driving force, with maximum value ${F_{\max }}$ and angular frequency ${\omega _d}$ , the amplitude of driven oscillations is,

$A = \frac{{{F_{\max }}}}{{\sqrt {{{\left( {k - m{\omega _d}^2} \right)}^2} + {{\left( {b{\omega _d}} \right)}^2}} }}$

Here, ${\omega _d}$ is the angular frequency if driven oscillation, $m$ is the mass, $k$ is the force constant and $b$ is the damping constant.

Here, ${F_{\max }} = nF$ where, $n$ is the number of people and $F$ is the force applied by each person.

Substitute $2000{\rm{ kg / m}}$ for $\lambda$ and $144{\rm{ m}}$ for $L$ in the equation $m = \lambda L$ .

$\begin{array}{c}\\m = \left( {2000{\rm{ kg / m}}} \right)\left( {144{\rm{ m}}} \right)\\\\ = 288000{\rm{ kg}}\\\end{array}$

The expression for the time period is,

$T = \frac{1}{\nu }$

Substitute $1{\rm{ Hz}}$ for $\nu$ in the above equation.

$\begin{array}{c}\\T = \frac{1}{{1{\rm{ Hz}}}}\\\\ = 1{\rm{ s}}\\\end{array}$

It is given that the undriven oscillations will decay to $\frac{1}{e}$ of its original value in a time $t = 6{\rm{ T}}$ .

Thus, the amplitude at time $t = 6{\rm{ T}}$ will be,

$A = {A_0}{e^{ - \frac{{bt}}{{2m}}}}$

Substitute $\frac{{{A_0}}}{e}$ for $A$ and $6T$ for $t$ in the above equation.

$\frac{{{A_0}}}{e} = {A_0}{e^{ - \frac{{b\left( {6T} \right)}}{{2m}}}}$

Rearrange the above equation for b.

$\begin{array}{c}\\\frac{{{A_0}}}{e} = {A_0}{e^{ - \frac{{b\left( {6T} \right)}}{{2m}}}}\\\\\frac{1}{e} = \frac{1}{{{e^{^{ - \frac{{b\left( {6T} \right)}}{{2m}}}}}}}\\\\1 = \frac{{6bT}}{{2m}}\\\\b = \frac{m}{{3T}}\\\end{array}$

Substitute 288,000 kg for m and 1 s for T.

$\begin{array}{c}\\b = \frac{{288,000\;{\rm{kg}}}}{{3\left( {1\;{\rm{s}}} \right)}}\\\\ = 96000{\rm{ kg / s}}\\\end{array}$

The amplitude of driven oscillation is given by,

$A = \frac{{{F_{\max }}}}{{\sqrt {{{\left( {k - m{\omega _d}^2} \right)}^2} + {{\left( {b{\omega _d}} \right)}^2}} }}$

Substitute $m{\omega _d}^2$ for $k$ in the above equation.

$A = \frac{{{F_{\max }}}}{{b{\omega _d}}}$

Substitute $nF$ for ${F_{\max }}$ in the above equation.

$A = \frac{{nF}}{{b{\omega _d}}}$

Rearrange the above equation.

$n = \frac{{Ab{\omega _d}}}{F}$

Substitute $1{\rm{ Hz}}$ for $\nu$ in the equation ${\omega _d} = 2\pi \nu$ .

$\begin{array}{c}\\{\omega _d} = 2\pi \left( {1{\rm{ Hz}}} \right)\\\\ = 2\pi \\\end{array}$

Substitute $2\pi$ for ${\omega _d}$ , $25{\rm{ N}}$ for $F$ , $75{\rm{ mm}}$ for $A$ and $96000{\rm{ kg / s}}$ for $b$ in the above equation.

$\begin{array}{c}\\n = \frac{{\left( {75{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)\left( {96000{\rm{ kg / s}}} \right)\left( {2\pi } \right)}}{{25{\rm{ N}}}}\\\\ = 1810\\\end{array}$

Substitute 3b for b in the equation ${A_{\max }} = \frac{{{F_{\max }}}}{{b{\omega _d}}}$ .

${A_{\max }} = \frac{{{F_{\max }}}}{{3b{\omega _d}}}$

Substitute $b{\omega _d}A$ for ${F_{\max }}$ and 75 mm for A in the above equation.

$\begin{array}{c}\\A = \frac{{\left( {75{\rm{ mm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{1000{\rm{ mm}}}}} \right)}}{3}\\\\ = \frac{{0.075\;{\rm{m}}}}{3}\\\\ = 0.025\;{\rm{m}}\\\end{array}$

Ans:

The number of people walking on the bridge is 1810.

The maximum amplitude of the driven oscillation is 0.025 m.

The number of people walking on the bridge is 1810.

The maximum amplitude of the driven oscillation is 0.025 m.