Can you cause a resonance disaster?

The equation of motion of the bridge is

$M\ddot x= -kx - mg(1+\frac{1}{4} \cos(\omega_0 t)$

where $x$ is the vertical displacement. Dividing by $M$ yields

$\ddot x= -\omega_0^2 x - \frac{m}{M} g(1+\frac{1}{4} \cos(\omega_0 t)$

Since this is a multiple-choice question, and the answers are well separated, I will use dimensional (units of measurement) arguments to get the solution. The quantity that causes the amplitude to increase is $g' = \frac{m}{4M} g = 2.45 \times 10^{-4} m/s^2$ . This has the unit of $m/s^2$ . The required amplitude $A=0.25m$ has the units of $m$ . The third parameter in the problem is the frequency $\omega_0$ that has units of $1/s$ . The only possible combination that includes all 3 parameters and gives a time (units of $s$ ) is

$t= A/g' \omega_0= 4.5 hours$

This way we miss a factor of 2 relative to the more exact solution, but the best guess is still $9hours$ . (The other possibility to get a time out of these parameters is simply taking $1/\omega_0$ , but that does not depend on the other two parameters and therefore cannot be the answer.)

For a real bridge the friction is NOT negligible and it causes a decay of the oscillations. If the decay time is much shorter than 9 hours (very likely), a single person cannot generate a significant oscillation amplitude. With 500 soldiers the time is 500 times shorter than the 9 hours, $65s$ . If the decay time is comparable or longer, the oscillations will become significant.

The total force on the bridge result to $F = F_g + F_p + F_K = -Mg - mg \left(1 + \frac{1}{4} \cos(\omega_0 t)\right) - K z = (M + m) \ddot z \approx M \ddot z$ We splite the position $z(t) = z_0 + u(t)$ of the bridge into a static position $z_0 = \langle z \rangle$ and a dynamic deflection $u(t)$ . The average force over one period results to zero, so that this determines the average position $\langle F \rangle = -(M + m) g - K z_0 = 0 \quad \Rightarrow \quad z_0 = - \frac{(M + m) g}{K} \approx \frac{M g}{K}$ For the deflection $u(t)$ we find the following equation of motion $\begin{aligned} - \frac{1}{4} m g \cos(\omega_0 t) - K u(t) &= M \ddot u(t) \\ \Rightarrow \qquad \ddot u(t) + \omega_0^2 u(t) = - f_0 \cos(\omega_0 t) \end{aligned}$ with the resonance frequency $\omega_0 = \sqrt{\dfrac{K}{M}}$ and the nomalized driving force $f_0 = \dfrac{m g}{4 M}$ .

To solve the differential equation, we assume a since oscillation with varying amplitude $u(t) = a(t) \sin(\omega_0 t)$ Inserting this in the differential equation results $\ddot u(t) + \omega_0^2 u(t) = \ddot a(t) \sin(\omega_0 t) + 2 \omega_0 \dot a(t) \cos(\omega_0 t) = - f_0 \cos(\omega_0 t)$ Therefore $\ddot a(t) = 0$ , so that the amplitude increases linear with time $a(t) = -\frac{f_0}{2 \omega_0} t$ For $a(t) = 0.25\,\text{m}$ , we can estimate the time to $t = \frac{2 \omega_0}{f_0} \cdot 0.25 \,\text{m} = \frac{2 \cdot 16 \cdot 800,000}{80 \cdot 9.81}\, \text{s} \approx 32,600 \,\text{s}= 9 \,\text{h}$