Answers · The physics underneath

How can you predict the shape of a plucked string a moment after you let go?

In shortWrite the starting corner as a sum of the string's allowed standing shapes, let each shape oscillate at its own rate, then add them back together.

The shape a moment after release is the same sum of allowed standing shapes as at release, with each term multiplied by its own cosine. You predict it by finding how much of each allowed shape the initial corner contains, advancing each amount through its own phase, and adding the results. The corner itself is not an allowed shape, so it has no single frequency; it is a mixture, and the mixture is fixed at the instant you let go.

y(x)=\sum_{n=1}^{\infty} A_n\sin(n\pi x),\qquad A_n=2\int_0^1 y(x)\sin(n\pi x)\,dx;\qquad y(x,t)=\sum_{n=1}^{\infty} A_n\sin(n\pi x)\cos(\omega_n t),\quad \omega_n=n\omega_1.

A worked pluck

Take a string of unit length plucked at its centre to height 0.5, so the starting shape is $y(x)=0.5-|x-0.5|$ . The recipe gives $A_n=4\sin(n\pi/2)/(n^2\pi^2)$ for this triangle. For $n=1$ : $4/\pi^2=0.4053$ . For $n=3$ : $-4/(9\pi^2)=-0.0450$ . For $n=5$ : $4/(25\pi^2)=0.0162$ . For $n=7$ : $-4/(49\pi^2)=-0.00827$ . Every even $n$ gives zero because $\sin(n\pi/2)=0$ , so the centre pluck contains only the odd shapes. The amounts are not fitted by eye: the integral in the recipe returns each one directly.

FIG 1Building the corner left by a centre pluck

the shape the pluck leftthe slowest shape alonewith the third addedwith the fifth added too

The straight-sided triangle and the sum of one, two and three odd terms. Three terms already track the sides closely and leave only the tip rounded.

The amounts fall quickly, each odd term roughly a ninth, then a twenty-fifth, then a forty-ninth of the first in size, which is why three or four terms already look like the corner. The straight sides are tracked closely and only the tip stays rounded. Three terms give a recognisable corner, four give a better one, and the limit is exact. A finite sum is not the corner, but the whole infinite sum is, and for a kink the convergence leaves no overshoot.

Why stepping the whole shape fails

The natural guess is to take the whole corner and step it forward as a single object. That has no single rate to step it at, because a corner is not one of the shapes the string can hold; it is a mixture of shapes whose rates are different. If you watch only the total displacement, the separate oscillations are hidden inside it, and the motion looks as if it needs a new rule at every instant. The way out is not a better stepping rule for the corner. It is to stop treating the corner as one thing.

The recipe works because two different allowed shapes multiply together and integrate to zero along the string. Where they agree in sign they add, where they disagree they subtract, and the two exactly cancel. So multiplying by $\sin(n\pi x)$ and integrating hears only the $n$ -th ingredient and is deaf to all the others. The doubling in front of the integral is only a normalisation: it makes the recipe return 1 when applied to the shape it is asking about. That is why the amount of each shape is a property of the pluck, not of the later motion.

T=\frac{2\pi}{\omega_1},\qquad t=\frac{T}{10}\Rightarrow \cos(\omega_1t)=\cos(\pi/5)=0.8090,\quad \cos(\omega_3t)=\cos(3\pi/5)=-0.3090,\quad \cos(\omega_5t)=\cos(\pi)=-1.

At a tenth of the fundamental period the first three terms have become $0.4053(0.8090)\sin\pi x=0.3279\sin\pi x$ , plus $0.0450(0.3090)\sin3\pi x=0.0139\sin3\pi x$ , minus $0.0162\sin5\pi x$ . The signs of the third and fifth contributions have reversed from release, which is why the sum is no longer the same corner. The first term has shrunk by about a fifth, while the short shapes have flipped, so the tip moves first and the sides follow.

FIG 2

n	A_n	A_n at t=T/10
1	0.4053	0.3279
3	-0.0450	0.0139
5	0.0162	-0.0162
7	-0.00827	0.00256

Where the method stops

Two assumptions are doing work. The string must be linear and uniform with fixed ends, and the release must have zero initial speed; a release with a push adds sine terms whose amounts come from the initial velocity. For a corner the infinite sum closes in from both sides with nothing left over. If the starting shape had a jump instead of a corner, finite sums would overshoot near the jump, and adding more terms would narrow the overshoot without shrinking it. A pluck leaves a kink, not a jump, so the method is safe here.

For the pluck, the arithmetic above is the whole prediction: the starting corner fixes the amounts once, the string fixes the rates once, and the shape at any later moment is the sum of those amounts turning at those rates. The pitch you hear is the slowest rate; the tone is the set of amounts. That is why plucking at the middle removes half the ladder and plucking near the bridge loads the short shapes. The same numbers would predict the shape after any other time, because the phases are the only things that change.

Worth keeping

A plucked string's starting corner is a sum of the string's allowed standing shapes, and the amounts are fixed at release.
Each allowed shape then oscillates at its own rate, so the later shape is the same sum with each term phase-shifted.
Plucking position changes the mixture, not the rates, which is why it changes tone but not pitch.

Where this goes next

This question is one step of Waves, Modes and Frequency, a course that works through its subject from the beginning. The page that teaches this part of it is Any Shape Is a Sum of Simple Ones, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that page The whole course