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I have a question regarding the formation of loops when a string is plucked at different fractional lengths.

In a book I referenced, it is stated that plucking a string at 1/6 of its length produces 3 loops. However, based on other sources I've read, the understanding is different.

From what I've gathered:

Plucking a string at half its length ($L/2$) produces 1 loop.

Plucking at $L/4$ produces 2 loops.

Therefore, for 3 loops, it seems logical that the length should be L/8, following a general formula of $L/2^n$, where $n$ represents the number of loops. For example:

To get 1 loop, $n=1$, so $L/2^1 = L/2$.

To get 2 loops, $n=2$, so $L/2^2 = L/4$.

To get 3 loops, $n=3$, so $L/2^3 = L/8$.

Can someone clarify if this reasoning is correct, or explain why a string plucked at 1/6 of its length would produce 3 loops?

Third mode of vibration

Sancol.
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Engr Fahad Safi
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1 Answers1

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One simple way to get some intuition is to think of it like this: After plucking, the string would like to relax into a sinusoidal wave which is zero at the edges and has a maximum amplitude (antinode) where it was plucked. In your case, the three loop wave has its antinodes at the points $x = L/6, L/2, 5L/6 $. This indeed is a sinusoidal wave with an antinode at the plucking spot.

Based on this, we would expect the following relation: A standing wave fits an integer number of nodes on the string, and these will be separated by $\lambda/2$ as in your illustration. To get $n$ loops you need to pluck an antinode of this wave. The first antinode will be at $x=\lambda/4$. For $n$ loops on a string of length $L$, the wavelength is given by $n \lambda/2 = L \Rightarrow \lambda = 2L/n$, and therefore the plucking point is at $x = L/(2n)$. Let us check this against the first three results:

  • $n=1, x = L/2$
  • $n=2, x = L/4$
  • $n=3, x = L/6$
  • Etc.

This is just the intuition, and not necessarily the full picture. Hopefully the book explains this in more technical detail, otherwise I would advise you to look around for deeper answers, see e.g. this answer.

Codename 47
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