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Hi I have trouble with tuplets. There is a tuplet that has different note values under the same beam ( 16 th. And 8th.) There are 4 16th notes and 1 8th note at the end. There is a number 6 under the beam. I think it is a sextuplet. I don't understand how the tuplet works and how I play it. Thanks.!

Here is a pic. It is the beam where there is a 6 under

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  • play it like a triplet where 2 of the 8th notes are divided in 16th. Sep 14 '19 at 20:42
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OK. You have not given us an image (which would really help) but lets try to analyse this.

So there are four 16th notes and one 8th note. The 8th note is worth two 16th notes isn't it. So you have a total of six 16th note lengths in all. Same length as three 8th notes.

So you are correct that it is a sextuplet, or to think of it another way, a triplet where two of the three notes have each been further subdivided into two notes.

The whole thing should add up to the length of a quarter note.

I am going to make the assumption that the beat is a quarter note: you need to divide this beat into three and further divide two of those three in half.

I know it sound complicated but its quite straightforward when you get it.

If it helps (and it might not, but I'll say it anyway) it is the same pattern as a bar of 3/4 where the first four notes are 8ths and the last one is a quarter.

Hope that helps

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just play them like triplets whereby two of the 8th notes are divided in 16th: instead of da-da-da you play da-daba-daba, (or daba-daba-da) ...

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  • But there isn't 2 8th noted there is only 1 8th note. But still that is some good advice. Thanks. Sep 14 '19 at 22:29
  • the triplets of 8th notes are three 8th notes: I say: there is only one 8th note and the two others are divided in four 16th notes. That means: two 16th notes are corresponding to one 8th note. daba stands four two 16th notes = > 1/8 (da) Sep 15 '19 at 7:23

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