Natural angular frequency of a bridge

[Linus Pauling]

1.

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

A=\frac{F_{\rm max}}{\sqrt{(k-m{\omega_d}^2)^2+(b\omega_d)^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 \rm N at a frequency of about 1 \rm Hz. Given that the mass of the bridge is about 2000 \rm kg per linear meter, how many people were walking along the 144-\rm m-long central span of the bridge at one time, when an oscillation amplitude of 75 \rm 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.3. F_max = A_res * b*omega_d

omega_d is the frequency of the driving force of a person, which is 2*pi*1Hz = 2pi
A_res is the resonance amplitude, b is the damping constant

#people = F_max / F_avg
= (A_res *b*omega_d) / F_avg

plugging in (m*omega_d)/6pi for b, which I already calculated, as well as plugging in A_res and F_avg, I obtain:

288*omega_n

How do I calculate the natural angular frequency of the bridge?

 

[Linus Pauling]

Is this understandable?

 

[kerimek]

To calculate the natural angular frequency of the bridge, we need to use the formula for natural frequency:

omega_n = sqrt(k/m)

Where k is the force constant of the system and m is the mass of the bridge.

In this case, we are given the mass of the bridge (2000 kg per linear meter) and the length of the central span (144 m). We can calculate the force constant by dividing the mass by the length:

k = m/L = 2000 kg/144 m = 13.89 kg/m

Now we can plug this into the formula for natural frequency:

omega_n = sqrt(13.89 kg/m / 2000 kg/m) = sqrt(0.006945) = 0.0833 rad/s

Therefore, the natural angular frequency of the bridge is 0.0833 rad/s.