Let's say we have a coffee cup as shown below.

The coffee cup

The cup is, in contrast to what's shown on the image, empty (I had some trouble finding free good-quality top-view coffee cup images). When we tap on $\frac14,\frac12,\frac34$-th of the coffee cup (measured from the "ear", not sure what that's called in English), showed with the red dots, the cup makes a lower sound then when we tap on $\frac 18,\frac38,\frac58,\frac78$th of the cup (shown by the blue dots). I have tested this on multiple types of cups and is thus definitely not an opinion-based question.

Why is this? Please make your answer as simple as possible, as I'm just a simple mathematician, and know nothing in particular about physics (although I have a basic understanding, of course).


2 Answers 2


The answer is related to the asymmetry of the cup.

The sound is, as we know, is resulting from the vibration of the cup. For a qualitative explanation it suffices to compare it with a vibrating disk, like the image below.

enter image description here
(source: psu.edu)

This shows only the main modes, which is the main one (the ones with higher amplitudes) in the infinite series, but the following reasoning applies to all of them.

The image shows a vibrating circle, but it would have similarities with the cup if you imagine the center pushed down. I will refer to the lines in the surface which do not bend as the "axes". We can see for this mode there are 2 axes.

Tapping on the red dots makes the cup's axes for this mode to lay along the blue dots. So unlike the our circle in the image, one of the vibrating quarters is unequal to the rest, it has the added mass of the ear. Hence it will vibrate slower than in the opposite case...

Tapping on the blue dots, the ear falls on one of the edges and almost does not move. The four quarters are equal and lighter than in the previous case, and move lightlier and hence produce a higher frequency.

  • $\begingroup$ Such an easily understandable answer, exactly what I was looking for. Thanks a lot! $\endgroup$
    – user108361
    Feb 15, 2016 at 15:35

If you look at the animation in the animation that @mhleo provided you will see that there a positions with no movement called nodes and positions where the amplitude of motion is maximum called antinodes.
The node to node distance is equal to half a wavelength of the sound wave in the cup and that is also true of the antinode to antinode distance.
The longer the wavelength the higher the frequency so if you double the wavelength you halve the frequency.

It is likely that the part of the cup with the handle acts as a node because of the increased mass in that region.
Where you tap the cup also probably produces a node.

Tapping on a red spot produces nodes at all the red spots with antinodes at all the blue spots and this produces a note of a particular frequency.

Tapping on a blue spot produces nodes at all the blue spots and also at the red spots because of the presence of the handle. So the node to node distance (red dot to blue dot) in this case is half that of the blue dot to blue dot distance mentioned before.

So the tapping on a blue dot frequency is higher than that of the tapping on a red dot frequency.

  • $\begingroup$ Thanks for your answer. I only miss one detail that I don't understand - could you explain why more nodes (or a smaller node distance) increases the vibration speed? $\endgroup$
    – user108361
    Feb 15, 2016 at 16:32
  • $\begingroup$ It is the vibrational frequency which is increased. wavelength $\times$ frequency = speed of wave. Decreasing the node to node distance decreases the wavelength and hence increases the frequency. Think of a vibrating violin string which has nodes at each end. What happens to the frequency of the note emitted is the string is made shorter so the nodes are closer together - the frequency of the note increases. $\endgroup$
    – Farcher
    Feb 15, 2016 at 16:51
  • $\begingroup$ So at the blue dots, there are 8 nodes, and at the red dots, there are 4 - this means that the node-to-node distance doubles, so should the sound then not be on entire octave higher? Because in practice, this is not the case $\endgroup$
    – user108361
    Feb 15, 2016 at 16:55
  • $\begingroup$ I have just outlined an idea. There is asymmetry with handle which might make a difference. The modes (patterns of nodes and antinodes) are not one dimensional they are at least two dimensional. It might be that the modes you excited where not exactly as I described. And after looking on the Internet as often is the case here is the answer from somebody who seems to know youtube.com/watch?v=oFRQkNMwrL0 I think that it is an excellent video and so is the next one about the vibrations in a guitar. $\endgroup$
    – Farcher
    Feb 15, 2016 at 17:01

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