## Mugging

After writing about ‘singing glasses’ previously, I was discussing the effect with my friend and colleague Andrew Hanson.

Have you done the mug thing?” he asked. And proceeded to hit the rim of a mug with a spoon.

As he moved the location at which he struck the mug, the pitch of the note changed.

And then he explained. Looking from the top, if the handle of the mug is at 12 O’clock, then:

• Striking the mug at 3, 6, 9, or 12 produces one note – a lower note.
• Striking the mug between 1 & 2, between 4 & 5, between 7 & 8 and between 10 & 11 produced a second note – a higher note.

Andrew said the explanation was that in a mug, there are two types of flexural oscillations, and the frequency of the oscillations depends on whether the handle moves or not.

I was fascinated. How had I never noticed that before? And why was the note that sounded when the mug was struck ‘on the quarter hours’ lower?

This is a very long article, and I apologise. But the physics of this phenomenon is complex and it took me a long time to get to the bottom of it.

Investigations

I picked 5 mugs from our domestic collection which were as straight-sided as possible, and which had walls which were as thin as possible. I thought these choices would make the vibrational spectrum simple.

First I measured the mugs: their diameter and the wall thickness, and then I picked a wooden striker and started hitting the mugs. (Speadsheet)

The place where one strikes the mug produces an oscillation with the striking location as a local maximum of the oscillation.

For a glass, it wouldn’t matter where one struck – any location on the rim is equivalent to any other. But for mugs, the striking position matters because of the handle.

To see if I could understand what was going on I arranged the mugs in size order, from the smallest radius to the largest and struck each one three times at each location.

You can hear the sound here.

I also recorded the ‘spectrogram’ using the wonderful Spectrum View app for the iPhone. A screenshot from the app is shown below together with the mugs which made the noise.

A spectrogram shows:

• time along the horizontal axis
• frequency along the vertical axis
• and the loudness of a sound at a particular frequency and time is shown by the colour: blue is quiet and yellow and red are loud.

On the spectrogram above one can see vertical lines which result from the ‘impulse’ sound of me hitting the mugs. This dies away quickly and one is just left with the ‘ringing’ of the mugs which I have outlined with dotted lines.

One can see that each mug rings at two closely-spaced frequencies. The two notes differ in frequency by between 5% and 15%.

Hitting the rim at either location produces mainly one mode of oscillation, but also a little of the other.

Let’s get numerical!

I used the app to locate the frequency of each note and plotted it on a graph of the frequency versus the mug diameter. I plotted each ringing note as a red dot, and their average as a black dot.

There was a general trend to lower frequency for the larger mugs, but the Toronto mug didn’t fit that trend.

I noticed that the walls of the Toronto mug were much thicker than the other mugs. So I wondered whether I could compensate for this by dividing the frequency by the thickness of the wall.

I did this based on the idea that the speed of the wave would be proportional to the  rigidity of the mug wall against bending. And that rigidity might be roughly proportional to the wall thickness. This seemed to be confirmed because the formerly ‘anomalous’ mug frequency now sat quite sweetly on a smooth trend.

Now that the data seemed to fit a trend, I felt I was getting a handle on this problem. Could I understand the dependence of the resonance frequency on diameter?

Frequency

All waves obey the wave formula v = f λ. That is, the speed of the wave, v, is the product of the frequency of the wave, f, and its wavelength  λ.

For the waves on these mugs, the wavelength of the wave which runs around the rim of the cup is just half the perimeter i.e.  λ = π D/ 2 where D is the diameter of the mug.

• So if all the waves travel with the same speed, then the resonance frequency should vary with diameter as f = 2 v / (π D) i.e. inversely proportional to the diameter.
• However, for flexural waves, the material supporting the wave becomes floppier at longer wavelengths, and the speed of a flexural wave should fall with increasing wavelength. If this were the case we would expect the resonance frequency to vary inversely as the diameter squared.

Which of the above cases described the data best? I have plotted the two predictions on the graph below.

I adjusted the speeds of the waves to match the data for large mugs and then calculated how it should vary for smaller diameter mugs.

Overall I think it is the theory in which the  speed of the flexural wave changes with wavelength that matches the data best.

Test

So now I had a theory that the speed of a flexural wave that runs around the rim of a mug is:

• Proportional to the wall thickness
• Inversely proportional to the diameter
• So the resonance frequencies are inversely proportional to the diameter squared.

After I had finished all these measurements I came across another mug that I could have included in the study – my wife’s ‘Do more of what makes you happy‘ mug.

I decided to see if I could predict its resonant frequency from measurements of its wall thickness and diameter.

I have plotted data on this mug as crosses (×) on the graphs above and I think that overall it fits the trends rather well.

The two notes

If you belong to the minority who have read this far, then well done. It is only you who get to understand the final point about why the ‘quarter hour’ notes are lower.

What is the role of the handle? There are two possibilities.

• Does it act as an extra mass which slows down the wave?
• Or does it provide extra stiffness, which would speed up the wave?

Since the quarter hour notes are lower, it seems that it is the extra mass which appears to dominate in the mugs I have examined.

### 3 Responses to “Mugging”

1. edhui Says:

Great post. You are now only one step away from following Feynman and getting interested in throat singing. Have you tried? I have. In the car. When nobody can hear. If you set up a background hum, and shape your mouth, you can definitely hear different harmonics of your base hum. I thought it was because bone transmission enhanced the harmonics, but the spectrograph in the iPad (later- not in the car!) showed the harmonics were clearly audible externally. Maybe my skull was ringing like a mug. This guy does it better, though. Ed

2. Gianluca Says:

Nice one, Michael! And beautifully explained!

As you know, I have been recently playing with the idea of applying theories from quantum mechanics (i.e. the theory that explains how atoms and molecules work) to acoustics (i.e. the science that explains how stuff sounds).

Since this seems appropriate to the idea behind your blog (protons and electrons everywhere!…now also in terms of the science that explains how they work) I wonder if one could use “symmetry” to explain what you have found.

“Simmetry” is a powerful tool in the world of atoms, let’s see what happens when we apply it here.

While a glass has the symmetry of a cylinder (i.e. you can go around it singing ring-a-ring-o’-roses and nothing changes), the mug has the handle that breaks the symmetry (i.w. while you sing and dance around the mug, you end up meeting it). Nothing new here.

Symmetry tells us that this splits the resonance of the mug in two: one oscillation at higher frequency and one at lower frequency. Just like happens to the molecule of CO2 that I remember from your lessons, that can stretch in two ways.

Now, which frequency is associated to which movement? In your drawings, you seem to imply that the two modes (the “quarter” mode and the “between quarters” mode) have the same wavelength, but since they have a different frequency, should they not have a different wavelength too?
In my “ring-a-ring-o’-roses” approach, I would need to miss a step to go around the handle. If the wavelength is linked how many times I can repeat the same movement around the circumference while dancing, then a lower number of steps means a larger wavelength and therefore a lower frequency.

Something to discuss with a cup of tea in front of us…

• protonsforbreakfast Says:

Ciao. Indeed.

The general approach to this problem as I have described it is an example of the grandly titled “First Order Perturbation Theory”. Here the basis states (aka ‘Normal Modes of vibration’) of the system are considered to be unchanged, and the effect of the perturbation (the handle) is estimated. Thus we assume that there is no change in the nature of the waves which run around the rim. The justification for this is that the change in the resonant frequencies is small – only about 5% to 10% of the frequency.

Of course we really need at least second order perturbation theory. Here we calculate the change in the normal modes caused by the perturbation.

In fact I think you can detect this change. For some of the mugs I was convinced that the best place to hit it was not at the quarter hour marks. I thought it was on the quarter hour mark for the 12 and the 6 (as expected by symmetry) but for the other quarter hour marks the place to stricke it had shifted away from the 9 and the 3.

Happy ‘pinging’.

M