explaining how piano notes decay depending on their frequencies.
To explain the graph, the x-axis is frequency (given in midi note index. To give you a sense of conversion, MIDI 57 = A3 (220Hz) and MIDI 69 = A4 (440hz)). The lower you go on the y-axis, the faster the decay.
Clearly, as you move right the data points start moving down (the higher the frequency the faster the decay). But there’s not a smooth curve; there’s a cloud of points. So, if you take any one point on the x-axis, let’s say MIDI 69. There is a spread of y values, i.e. different decay rates. There’s also a spread of grayscale, which means that this frequency will decay at lots of different rates depending on the how high order a partial it is!
I know it’s not a terribly satisfactory answer, but the underlying physics is sufficiently complex that a satisfactory answer doesn’t quite exist. The gist is: usually higher frequencies decay faster, but not always, and there are complicated reasons why!
The higher the string, the shorter and thinner. The lower the string, the longer and thicker. The lower strings have more mass and do not let go of the vibrations as quickly as the higher strings. Additionally, the lower strings have more harmonics and more opportunity to resonate with other strings in the piano, which adds to the sustain. If you need more detail, you will probably need to find it in a physics text.
Aside from the increasing momentum of lower pitch strings, note that the damping force is practically the same for all strings during free vibration. So the loss rate of energy is the same for any string. This makes lower pitch strings take longer time to consume their energy.
One can design a special instrument that gradually increases the damping force as the note goes lower so that the sustain times become equal. However in this special case, still the damping force of the resonating amplifier body (for example the wood board of a guitar or piano, as well as the room around the instrument) would stay the same and provide slightly longer sustain for lower notes.
You can test this phenomenon on any string instrument. Just play a bass note and then stop it using your hands/bow, and repeat the same for a high note. You will hear that the bass note resonating longer on the board.
Also note that pianos have bigger hammers and dampers for the lower notes for the same reason. You need to generate more energy and then consume it back.
Another example is, on a piano you can hear low notes sustaining longer than high notes when you lift your finger from the key.
Let’s consider a wire strung between two posts. After you hit it with a hammer two waves pulses propagate up and down the string, one in each direction. They hit the end posts, bounce in the other direction, and so on; two pulses racing back and forth along the length of the string.
The fundamental frequency, ie the pitch of the string, is the inverse of of the round trip time.
First, consider the case where the internal friction of the string is very low. Then the losses occur when the pluses hit the end posts. For higher frequency vibrations, this happens more rapidly. Assuming that the same fraction of the wave energy is lost from the string (and transferred to the sound board) each time, you’d expect the sound to dissipate more rapidly for higher frequencies.
Similar considerations apply when thinking about internal losses — a given short segment of the string flexes and then flattens as the pulse moves through it. And again, if each flex dissipates some energy, then more energy is lost per unit time for higher frequency strings. But this gets confounded by the fact that there are more such short segments in a longer string. A naive application of this logic leads one to conclude that this might result in a frequency-independent term of the losses. (From there you can think about whether the internal losses depend on the rate of change of the shape...)
So, to first order, keeping everything other than the length of the string fixed, you’d expect that higher pitched strings would lose energy more 