Why did this happen to the Millenium Bridge in London on the day that it opened?
When a horizontal, frictionless oscillator is subject to linear drag, the resulting equation of motion is mx¨+bx˙+kx=0.
a) Verify that the above equation is true, then simplify it using β=2mb and ω0=mk.
b) Try the solution x(t)=ert then solve for r in terms of β and ω0.
c) Interpret the solutions for β<ω0 and β>ω0.
a) If we suppose that the cart is to the right of the equilibrium point (i.e. in the +x-direction), then according to the free body diagram
b) Let's try the solution x(t)=ert.
Therefore the equation from Newton's Second Law becomes
We can solve this for r.
c) Since r has two different values, that means that our general solution can be written as a linear sum of those two solutions.
If β<ω0, then the square roots become imaginary.
This means the terms in parentheses represent oscillations with frequency ω1=ω02−β2. The stuff out front of the parentheses, meanwhile, represent an exponential decay. Putting it all together, when β<ω0 the result is an exponentially decaying oscillator.
If β>ω0, then the square root is positive, which means that no oscillation occurs.
In the simulation below, try playing around with the parameters so that you can see what each of these situations looks like.
from__future__importdivision,print_functionfromvpythonimport*importmatplotlib.pyplotasplt%matplotlibinlineL=1#relaxed length of springomega0=10#natural frequencybeta=15#decay parameterx0=L+0.2#starting displacementt=0dt=0.01#time incrementxdata=#set up data array for plotting x-coordinatestdata=#set up data array for plotting t-coordinates#draw box and springbox=box(pos=vec(x0,0.2,0),length=0.4,height=0.4,width=0.4,color=color.blue,v=vec(0,0,0))spring=helix(pos=vec(0,0.2,0),axis=box.pos-vec(0,0.2,0),radius=0.1)whilet<=10:rate(1/dt)box.a=-2*beta*box.v-omega0**2*(box.pos-vec(L,0.2,0))#calculate accelerationbox.v=box.v+box.a*dt#update velocitybox.pos=box.pos+box.v*dt#update positionspring.axis=box.pos-vec(0,0.2,0)#change spring lengtht=t+dt#increment timexdata.append(box.pos.x)#add x-coordinate to data array for plotting purposestdata.append(t)#add t to data array for plotting purposesplt.xlabel("t (s)")plt.ylabel("x (m)")plt.plot(tdata,xdata,"b-")
[<matplotlib.lines.Line2D at 0x7f6651917780>]
In the previous problem, we didn't look at the case β=ω0.
a) Show that the general solution x(t)=e−βt(C1eβ2−ω02t+C2e−β2−ω02t) is not able to describe the cart if the initial conditions are such that x(0)=0 with x˙(0)=v0=0.
b) Show that x(t)=C1e−βt+C2te−βt is a solution to x¨+2βx˙+ω02x=0 when β=ω0.
c) Solve for the unknown coefficients in the special case x(0)=0 and x˙(0)=v0.
According to the first equation, C1=−C2. But if that is true, then the second equation is immediately violated.
in damped oscillators, the amplitude of oscillations decreases as energy is dissipated away
in the special case of linear drag, the equation of motion is x¨+2βx˙+ω02x=0, where β=2mb is a decay parameter and ω0=mk is the natural frequency (i.e. the oscillation frequency in the absence of damping)
how does β compare to ω0?
Linear Differential Operators
Suppose we define D=dt2d2+2βdtd+ω02.
Then the linear damped oscillator equation can be written as Dx=0.
Now suppose that there is an additional external force that doesn't depend on x(t) at all: x¨+2βx˙+ω02x=f(t), where f(t)≡mF(t). (This is called a forced oscillator.) Then the linear damped oscillator equation could be written as Dx=f.
We will now show that the full solution to Dx=f consists of the solution to Dx=fas well as the solution to Dx=0.
First, if D is a linear operator, then D(ax1+bx2)=aDx1+bDx2. Suppose we somehow found a solution that works in the forced oscillator equation. We will call this the particular solution.
We will call the solution to the unforced oscillator the homogeneous solution.
Now because D is a linear operator, then D(xp+xh)=Dxp+Dxh=f+0=f.
Therefore the most general complete solution to Dx=f must also include solutions to the corresponding unforced (i.e. homogeneous) oscillator equation.
An oscillator is forced with a forcing function f(t)=f0cos(ωt), where ω=ω0. (An example similar to this would be pushing someone who is on a swing.)
Show that x(t)=Acos(ωt−δ) satisfies the equation, where A2=(ω02−ω2)2+4β2ω2f02 and δ=arctan(ω02−ω22βω).
First it may help to write x(t)=ℜ[Aei(ωt−δ)].
Then the forced oscillator equation is ℜ[−ω2Aei(ωt−δ)+2βωiAei(ωt−δ)+ω02Aei(ωt−δ)=f0eiωt].
Every term contains eiωt, so we can cancel it out.
Then we can match the real terms and the imaginary terms.
To solve for δ we can divide the first equation by the second.
The general solution to x¨+2βx˙+ω02=f0cos(ωt) is x(t)=Acos(ωt−δ)+e−βt(C1eβ2−ω02t+C2e−β2−ω02t), where A2=(ω02−ω2)2+4β2ω2f02 and δ=arctan(ω02−ω22βω).
a) Describe the long-term behavior of a damped, sinusoidally forced oscillator.
b) For what value of the forcing frequency ω is the amplitude of the long-term oscillations the greatest?
a) As time goes on, the e−βt term makes the natural oscillation disappear. It is because of this that the natural oscillation terms are referred to as transients. What remains is the oscillation due to the forcing.
b) A is maximized when ω=ω0. Recall that ω0 is the frequency of oscillation without forcing. This means that the amplitude of oscillation is biggest when the driving frequency is equal to the natural frequency. This condition is known as resonance. The idea of resonance makes sense when you think about pushing someone on a swing. You push only when they come back to you. You wouldn't give them a starting push and then push them again while they are on their way back to you mid-swing.
This, by the way, helps us answer what happened to the Millenium Bridge. By an unfortunate accident, the tension cables of the bridge happened to create a system in which the natural frequency matched the frequency of footfalls on the bridge on that day. It didn't help that the swinging made people adjust their gait in such a way that amplified the problem.
So how was the problem solved? The bridge was closed down and extra dampeners were added to the bridge.
The equation of motion of a damped, forced oscillator is x¨+2βx˙+ω02x=f0cos(ωt) with x(0)=0, x˙(0)=0, ω=2π, ω0=5ω, β=ω0/20, and f0=1000. Recall that the general solution is