3
$\begingroup$

I just read this answer by rob on the question of why harmonics occur when plucking the string, in the answer I have a doubt on this point:

It's worth pointing out that you have some control over which harmonics you excite by choosing where you pluck the string.

This is an intuitive thing as otherwise I can't imagine music being played from instruments but I can't exactly pin down a reason for why we have control over the harmonics excited. Hence, I seek an answer which can give an exact reason for the above.

$\endgroup$
1
  • 1
    $\begingroup$ See also "natural" and "artificial" harmonics as defined in the music community, which may give you some more insight as to how standing waves depend on the location of nodes and antinodes. $\endgroup$ Apr 7, 2021 at 14:13

2 Answers 2

2
$\begingroup$

If you solve the wave equation, you will find that the general solution can be written as a linear superposition of the normal modes or harmonics $\sin(k_n x)$ as:

$$y(x,t) = \sum_{n} c_n \sin(k_n x) \cos(\omega_n t),$$

This solution is for the case when the string has some initial displacement (i.e., it is plucked) at $t=0$, and $k_n = n \pi/L$.

Now, the $c_n$s are the "weights" of the different harmonics in the final solution. However, as you can see, they are independent of $t$, and so a different set of $c_n$s would represent a different solution on the string. So how would we normally find the different $c_n$s? Well, we'd use the initial conditions ("How the string was plucked") to do this, since if the string was plucked with some profile $y_0(x)$ at $t=0$, then we would require that: $$y_0(x) = \sum_n c_n \sin\left( \frac{n\pi x}{L}\right) \quad \quad \implies \quad \quad c_n = \frac{1}{L}\int_0^L y_0(x) \sin\left( \frac{n\pi x}{L}\right) \text{d}x.$$

Thus in general, different forms of $y_0(x)$ give different $c_n$s, meaning a different set of excited harmonics. So it's not just where the string was plucked that would excite different harmonics, but also how it was plucked (the "shape" of the initial displacement).


If you want a simple "intuitive" explanation, imagine that you could pluck a string in such a way that you created a node at $x = L/2$. (This can be done quite easily on most guitars.) Since the final solution must be a linear combination of the harmonics, it cannot contain any that do not have a node at $x=L/2$, since they would all have non-zero amplitudes there. As a result, all harmonics with an even number of nodes are eliminated from the solution (i.e., $c_0 = c_2 = c_4 =\dots= 0.$ A similar argument can be made for a string plucked at $x=L/3$, and so on.

$\endgroup$
1
  • 1
    $\begingroup$ Exactly right. Of course, if you pluck at some really ugly location (ratio of upper and lower string lengths implies a very high harmonic), the energy will quickly move to a collection of lower harmonics. $\endgroup$ Apr 7, 2021 at 14:11
2
$\begingroup$

The (half-)wave length of the oscillations excited in a string is commensurate with (i.e., a fraction of) the string length, which for the main harmonics is much bigger than the width of the finger. Thus, if the perturbation of the string by the finger is expanded in Fourier series in terms of the available modes/harmonics, it will necessarily contain many non-zero terms (the extreme version of this situation is white noise: a delta-function-shaped perturbation has equal spectral power on all frequencies - here however we have quantized (discrete) harmonics).

The power transferred to different harmonics depends on the amplitude of their oscillations at the point where the string is plucked. If we pluck the string at the node of the Harmonic, it is not excited at all (Note that the main harmonic is always excited, as its nodes are at the ends of the string.) On the other hand, if we pluck the harmonic near its maximum, it will have a lot of energy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.