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Are lower key harmonicas (G and A, for example) technically harder to bend than the ones with a higher key, such as E or D? If so, do I need to use some other technique on lower key harmonicas in order to achieve a nice sounding bend?

I'm asking this because last week I bought an harmonica in the key of A. However, I'm finding it much harder to bend (and play, actually) on the lower holes (between 0 and 6) than my other harmonicas (E, D and F - all of the same good quality).

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2 Answers 2

What about an engineer who is into music : )

Lower pitched notes require more air on the harmonia. The "length" of their vibration is also larger (so lower frequency).

This means you will be inhaling a greater voume of air to achieve the pressure difference required to reach the bend. This will usually be particularly noticeable on the first hole.

Take a look at the bending techniques described here, that might help you. Also, is the lower tuned harmonica of the same brand? As there might be some differences in the reeds used and a lower quality harmonica might be more difficult to bend correctly.

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The absolute change in frequency is smaller for lower pitches, not larger. eg 0.97 Hz between C0 and C#0, 249 Hz between C8 and C#8. Not that it matters, the proportional difference is constant for all notes so it is a null point and should be removed from the answer. –  Fergus Feb 19 '14 at 3:52
You are right. I was referring to the wavelength, not the frequency. I have adapted the answer. –  dorien Feb 19 '14 at 8:27
For a fixed wave speed such as sound in air, wavelength is inversely proportional to frequency so my initial comment still applies... –  Fergus Feb 19 '14 at 8:49
Ok, I see what you mean. Within one timeframe the change in frequency is the same, but the change for one "wave" will be bigger for lower notes, which is what got me confused. So because you look at a fixed timeframe, this doesn't matter. Will remove the sentence. –  dorien Feb 19 '14 at 8:57

This question is more suited to physics. stack exchange but here is a basic answer anyway...

Lower notes correspond to greater vibrating mass. Newtons second laws tells us tells us that force is proportional to mass for a given acceleration.

Force=mass x acceleration

So a greater force is required to vibrate a lower pitched reed. However, there are many variables in the case of the harmonica,: reed geometry, chamber geometry, reed material (elasticity, density, resonance characteristics etc), flow rate and it's relation to apparent force required, apparent volume, and most importantly how these variables change through the instruments range and during bending.

I have left a lot out, of you want more detailed answers try the physics stack exchange

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damn you science!!! –  Alexander Troup Feb 14 '14 at 20:56
Thank you for your answer, I really appreciate it. I'm not looking for a completely theoretical answer, though, so this is why I think music.SE is a better place to ask this instead of physics.SE. –  Streppel Feb 14 '14 at 21:23

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