I study about waves and I learn that the velocity of a transverse wave in a string depends only on the tension and the density of the string. The frequency depends only on the source (whatever generates the wave). The wavelength depends on both of them in the equation $c = f \lambda$ with $f$ the frequency and $\lambda$ the wavelength. But I have a question about string in guitar, the sound that the string make depends only on the frequency right? And if I tune the guitar then I enlarge the tension so the velocity gets bigger then why the sound is different?

Another question is, in standing wave if the frequency has not the right value of resonance (the first harmonic,etc) then it isn't count a standing wave even though 2 edges of a string attached to a wall? Or just the standing wave isn't full constructive?


1 Answer 1


Most probably, what you are referring to here is the equation

$$ f_{n} = \frac{n}{2 L} c = \frac{n}{2 L} \sqrt{\frac{T}{\mu}}, ~~~~~ n = 1, 2, \ldots $$

with $c$ being the speed of sound in the string, $L$ the total length of the string $T$ the tension of the string and $\mu$ its linear density. Number $n$ is just the harmonic index with $n = 1$ being the fundamental. Note that the speed of sound in the string in this linearised approach is given by $c = \sqrt{\frac{T}{\mu}}$.

When you tune the guitar string you alter its tension. By direct inspection of the equation above you'll see that the frequency of the fundamental (and all the harmonics since they are dependent on the fundamental) will change as $f \sim \sqrt{T}$.

More intuition about the resulting changes in speed along with the frequency can be built by inspecting the equation above and the fundamental wave equation that relates speed and frequency

$$c = \lambda f$$

with $\lambda$ being the wavelength. By changing the left hand side of this equation the right hand side must also change for the equality to hold. From the equation above you'll notice that the two quantities connected are frequency and speed of sound, so by changing the speed of sound you effectively alter the frequency of the right hand side of the second equation, leaving the wavelength unchanged. This makes sense, since for a standing wave to exist there must be a specific relation between the wavelength and the length of the string.

The frequencies supported by a string are very specific and depend on the dimensions of the system (string) and the boundary conditions (how is the string attached at the ends). For the equation presented above, the boundary conditions are $x \left( 0 \right) = x \left( L \right) = 0$ which means that the sting is not allowed to move on its edges. If these conditions are changed the allowed wavelengths will be different, thus changing the allowed frequencies.

If we consider solutions of the wave equation for this system of the form (corrected after Gert's comment)

$$ u \left( x, t \right) = \sum_{n = 1}^{\infty} A_{n} \cos \left( \frac{n \pi c t}{L} \right) \sin \left( \frac{n \pi x}{L} \right) $$

use the boundary conditions presented above and combine with the second equation you end up with the solutions for the allowed frequencies of the form given by the first equation (you have to use $c = \sqrt{\frac{T}{\mu}}$ to epxress the speed of sound). Please note that this is a standing wave solution derived from two waves traveling in opposite directions.

As you can, see, there are only specific frequencies allowed in the system based on the boundary conditions and the dimensions of the system. You use the term resonance here, which is not exactly equal to the topic of the natural frequencies of a system. What we have touched upon here is the natural frequencies of the system, which will be excited when the system is free of external forces and left to vibrate freely.

On the contrary, a resonance frequency, which is defined as the frequency for which an infinitesimal change will decrease the excitation amplitude (whether this is of amplitude, velocity, acceleration, pressure or whatever other quantity) of the system, refers to forced oscillations with some "source" providing the energy to sustain the vibration. In this case, the frequency the system will oscillate is always the frequency of the source (for a linear system at least). If this frequency coincides with one of the natural frequencies of the system (most probably some small corrections have to be made but those given above are adequate for a first-stage approximation) then we can talk about "hitting" a resonance.

Just to make it clear, when the system oscillates freely, it will do so in the frequencies given by the first equation. When the system is excited it will oscillate with the frequency provided by the source. If the latter happens to coincide with one of the natural frequencies, the amplitude (or velocity, or acceleration) will be maximised.

  • $\begingroup$ thanks for your reply. "when the system oscillates freely, it will do so in the frequencies given by the first equation" i don't understand that because if a person generates a wave through a string then it will oscillates with the frequency that the person provided , no?.I understand the relation between frequency and speed of sound in string by the equation but in the book i learn from "Fundamentals of physics by Halliday and Resnick" they wrote this segment ibb.co/2gTsVJG and because of that i got confused.So the frequency is not related to the velocity in excited systems?@ZaellixA $\endgroup$
    – Lourdes
    Commented Mar 27, 2022 at 9:44
  • $\begingroup$ This segment states two things. First of all, the second part of it declares that they are talking about forced oscillations (since they talk about external excitation). The first part states that the speed is constant for all frequencies in the string and depends only on the characteristics of the string (linear density and tension). This means that the medium is not dispersive (all frequencies have the same speed). As you can see in the first equation, the "allowed" frequencies for a system do depend on the speed which, as stated in the answer makes sense because you have to (cont.) $\endgroup$
    – ZaellixA
    Commented Mar 27, 2022 at 9:56
  • $\begingroup$ change the right hand side if the left hand side of equation $c = \lambda f$ for the equality to hold. So in the image you linked to, in the first question the speed does not change when you change the frequency since the speed is the same for all frequencies. From the above equation, if the left hand side does not change but $f$ changes, you have to change accordingly the wavelength $\lambda$ to counter the increase in frequency so that $c$ remains constant. Now, in the second question, if you change the tension you do change the speed so the right hand side of the equation (cont.) $\endgroup$
    – ZaellixA
    Commented Mar 27, 2022 at 9:59
  • 1
    $\begingroup$ "If we consider solutions of the wave equation for this system of the form" Instead of your presented solution it should be: $u(x,t)=\displaystyle\sum_{n=1}^{\infty}A_n\cos\Big(\frac{n\pi ct}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$ The $A_n$ form the amplitude spectrum, resposible for the timbre of the note. (sciencemadness.org/talk/…, which provides my solution to the wave equation of a guitar string) $\endgroup$
    – Gert
    Commented Mar 27, 2022 at 12:31
  • 1
    $\begingroup$ @ZaellixA you're welcome. $\endgroup$
    – Gert
    Commented Mar 27, 2022 at 22:33

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