Question

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at a frequency of about 1 HzHz. Given that the mass of the bridge is about 2000 kgkg per linear meter, how many people were walking along the 144-mm-long central span of the bridge at one time, when an oscillation amplitude of 75 mmmm was observed in that section of the bridge

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

   The amplitude of the lateral  force is  $F = 25 \ N$

   The frequency is   $f = 1 \ Hz$

   The mass of the bridge per unit length is  $\mu = 2000 \ kg /m$

    The length of the central span is  $d = 144 m$

     The oscillation amplitude of the section  considered at the time considered is  $A = 75 \ mm = 0.075 \ m$

      The time taken for the undriven oscillation to decay to $\frac{1}{e}$  of its original value is  t = 6T

Generally the mass of the section considered is mathematically represented as

            $m = \mu * d$

=>        $m = 2000 * 144$

=>        $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a  time period  t is mathematically represented as

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

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question  

               $\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=>            $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=>          $e^{-1} =e^{-\frac{b6T}{2m} }$

=>           $-\frac{3T b}{m} = -1$

=>         $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

           $A = \frac{n * F }{ b * 2 \pi }$

=>       $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=>       $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=>       $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=>       $n = 1810 \ people$