Why are strings of different thicknesses and materials used in a sitar or a violin? I guessed that to produce different frequency, different thicknesses and materials are used. Is this correct?


I guessed that to produce different frequency different thicknesses and materials are used , is it correct?

You're not too far off, though there's a bit more to it than that.

The speed of a wave on a string is given by $$ v = \sqrt{\frac{T}{\lambda}}, $$ where $T$ is the tension in the string (higher tension = more "tight") and $\lambda$ is the mass per length of the string. Moreover, if we have a length $L$ of string with the above tension and mass density, it will vibrate at a "natural" frequency of $$ f = \frac{v}{2L}. $$ (It also vibrates at integer multiples of this frequency, in the so-called "overtone series". More on this below.)

Putting these factors together, we can see that to get a high frequency (which corresponds to high pitch) for a given length, we can do one of two things: increase the tension, or decrease the mass. When you adjust the tuning pegs in the pegbox of a violin, you're changing the tension, which is why the pitch changes.

In principle, you could use the same material for all of the strings of a violin, and simply adjust them to different tensions. However, in practice, one wants to have all the tensions approximately the same; this means that the forces on the pegbox are more balanced; also, as pointed out by niels nielsen, it is harder to press down on a tighter string. The frequency of an open E string on a violin is about 3.4 times higher ($(\frac{3}{2})^3$ times, to be exact) than the frequency of an open G string; to produce the same frequency by adjusting the tension alone would require about $3.4^2 \approx 11$ times more tension in the higher string. This is impracticable, so in practice lighter strings have to be used for the higher strings, along with adjusting the tension between the strings.

As far as different materials go, this has more to do with controlling the timbre of the sound. The equation that I wrote up top for the speed of the wave is only strictly valid if the string has no resistance to bending. But thicker strings tend to be harder to bend, and steel strings tend to be harder to bend than gut or synthetic strings. This means that if the string is bending more, there's an additional force trying to "unbend" it; which effectively acts like increasing the tension. The net result is that the higher frequencies in the overtone series get shifted upwards, which affects one's overall perception of the quality of the sound created. It's a hugely complicated subject, and I'm only scratching the surface of it here; you can read more details at this answer.


Fundamental frequency of a stretched string, $v=\dfrac{1}{2L}\sqrt{\dfrac{T}{m}}$

When we use strings of different thicknesses and materials, they have different values of mass per unit length ($m$).So the strings will produce notes of different frequencies.


Yes, that is correct- each stringed instrument has its own unique scale length and pitch range. to obtain the correct pitch range for a given string length requires the right combination of string thickness, material, and tension force, which then determines how much strength the instrument needs to withstand the tension force, summed across all the strings.

In addition to all these considerations, the playability of the instrument has to be taken into account as well. This means for example that the strings cannot be so tight that they cannot be fingered properly by a human hand. It is a complicated business but one that furnishes music, which makes it worth while.


protected by Qmechanic Mar 20 '18 at 21:52

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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