You've heard the story of the singer who can shatter a glass just by singing a particular note real loudly. Well that may or may not be true, but it's an example of resonance. When you hear annoying buzzing sounds in your car, it is often something ``resonating" with the car engine. When a child, or adult, is on a swing, they can increase their amplitude of oscillation by ``pumping" the swing at just the right moments. That's also an example of resonance.

But resonance is even more common than that. Atoms is materials especially one with colors, resonate when light hits them. Light by the way applies a high frequency periodic force to electrons in a material. A microwave oven works on a similar principle. The microwaves apply periodic forces to water molecules causing them to resonate, and hence heat up. You might have seen some spectacular footage of some huge bridge collapsing right after it was built. That was an example of resonance. The wind was applying forces at just the right frequency to get the whole thing to swing wildly back and forth, ultimately collapsing.

So let's try to understand resonance. What you've got is a mass on a spring with some friction. You apply some periodic force at some frequency, and ask how much the system moves. If you apply it at just the right frequency and the damping isn't too big, you'd expect a small force would give a large response. Let's try to write down the equation we want to solve. $F_{net}$ now includes, the spring $-kx$, damping $-bv$, and an external force $F_{ext} ~=~ F_0 \cos(\omega t +\theta)$. We've added in a phase shift of $\theta$ because the applied force is not necessarily in phase with the harmonic motion. We'll talk more about that below. So

F_{net} ~=~ -kx -bv +F_0 \cos(\omega t +\theta)
\end{displaymath} (1.78)

which equals $ma$ so
ma + bv +kx ~=~ F_0 \cos(\omega t + \theta)
\end{displaymath} (1.79)

Hmm, this looks even worse than the damped harmonic oscillator because now the right hand side isn't zero! Well fortunately there's a simple way of figuring this out without too much algebra. The hard way would be to try a solution of the form

x(t) ~=~ A\cos(\omega t)
\end{displaymath} (1.80)

and plug it into $F_{net} ~=~ ma$ and solve for $A$. That is a bit painful so let's do it the easier way. It involves the same trick we used earlier, the two dimensional the problem:

m{\bf a}+ b{\bf v}+k{\bf r}~=~ {\bf F}(t)
\end{displaymath} (1.81)

What does this problem look like? Well we've already seen that it involves a bungey cord with ${\bf F}_{bungey} ~=~ -k{\bf r}$. But now we also have friction $F_{friction} ~=~ -b{\bf v}$. And now we have an external force ${\bf F}$. You can think of this external force as being applied to the object in some direction and the whole thing rotating around in a circle.

This is a lot like jetskiing in water with a bungey cord attached to a buoy.


The external force is applied by the motor of the jetski and if it's pointing in just the right direction, it'll go round the buoy in a circular path. The angle between the jetski and the radius vector stays constant at $\theta$.

We remember that if you consider just the x-component of the two dimensional problem, you should get the 1d problem. Let's check that it works. If we consider the x component of $m{\bf a}+ b{\bf v}+k{\bf r}$ you get $mx +bv_x+kx$. That looks OK. How about ${\bf F}$? The angle that this makes with the x axis is $\omega t + \theta$, so you get an x component of $F \cos(\omega t +\theta)$. That is the same as $F_0 \cos(\omega t + \theta)$ if you just make $F_0$ the magnitude of ${\bf F}$. So this two dimensional equation has as its x component the one dimensional equation we want to solve. The radius vector has an x component $x ~=~ R\cos(\omega t)$. So $R ~=~ A$ is the amplitude of oscillation of the 1d problem. This is all similar to what we did before for simple harmonic motion in section 1.1.5.

Let's try to figure out how the radius of the circle $R$, is related to the force that's applied and the angular velocity. Again we're applying ${\bf F}_{net} ~=~ m{\bf a}$. Now the free body diagram looks like


So we have

{\bf F}_{net} ~=~ -kR{\hat i}-b\omega R {\hat j}+ {\bf F}~=~ -m\omega^2 R{\hat i}
\end{displaymath} (1.82)

Here we've used $v~=~\omega R$ and $a ~=~ \omega^2 R$. Solving for ${\bf F}$

{\bf F}~=~ (kR-m\omega^2 R){\hat i}+ b\omega R{\hat j}
~=~ R \large ((k-m\omega^2 ){\hat i}+ b\omega {\hat j}\large )
\end{displaymath} (1.83)

This tells us the direction and magnitude of the force necessary to keep the jetski going around the circle. It has a magnitude

F ~=~ F_0 ~=~ R \sqrt{(k-m\omega^2 )^2 + b\omega }
\end{displaymath} (1.84)


A ~=~ R ~=~ {F_0 \over \sqrt{(k-m\omega^2 )^2 + (b\omega)^2 }}
\end{displaymath} (1.85)


\theta ~=~ \tan^{-1}\large ({ b\omega\over k-m\omega^2} \large )
\end{displaymath} (1.86)

Good, we've now understood the bungey cord - jetski problem, let's try to relate it back to the one dimensional resonance problem.

This says that if you apply a force with amplitude $F_0$, that the amplitude of resonance will just be as given above. Let's plot how this looks, taking $F_0 ~=~ k~=~ 1$ and $b^2 ~=~ 0.1$.


You see that the amplitude has a strong maximum close to $\omega ~=~ \sqrt{k/m}$. Frequencies close to that are at resonance. The smaller the damping $b$ is, the stronger the maximum, and the sharper the peak.

Now what about this mysterious angle $\theta$? This says that if you apply a force, the response to it, $x(t)$ will not be instantaneous by will be shifted in phase by an angle $\theta$ as shown below.


Close to resonance when $\omega ~=~ \sqrt{k/m}$, you can see that $\theta ~=~ \pi/2$. There is a 90 degree phase shift between the applied force and the displacement.

josh 2010-01-05