Question

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. The combined forces exerted by the pedestrians as they were walking in synchronization had a frequency very close to the natural frequency of the bridge, and so resonance occurred.
Consider an oscillating system of mass m and natural angular frequency ωn. When the system is subjected to a periodic external (driving) force, whose maximum value is Fmax and angular frequency is ωd, the amplitude of the driven oscillations is A=Fmax(k−mωd2)2+(bωd)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.
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 1

Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

$A(t)=A_oe^{\frac{-bt}{2m}}$

or amplitude for unforce oscillation

$\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}$

or

$\frac{6bt}{2m}=1$

or

$b=\frac{m}{3T}$

Now, provided in the question Amplitude of the driven oscillation

$A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}$

the value of the maximum amplitude is obtained $(k=m\omega_d^2)$

thus,

$A=\frac{F_{max}}{(b\omega_d}$

Now, for n people on the bridge

Fmax = nF

thus,

max amplitude

$0.075=\frac{nF}{((\frac{m}{3T})2\pi}$

or

n = 1810

hence, there were 1810 people on the bridge

b)$A=\frac{F_{max}}{(b\omega_d}$

since the effect of damping in the millenium bridge is 3 times

thus,

b=3b

therefore,

$A=\frac{F_{max}}{(3b\omega_d}$

or

$A=\frac{1}{(3}A_o$

or

$A=\frac{1}{(3}0.075$

or

A = 0.025 m = 25 mm