Shortening a guitar string to one-third its initial length will change its natural frequency by what factor?
A guitar string, when stretched within its elastic limits, obeys the wave equation. A little analysis of the motion of elastic strings gives the equation:
v = sqrt(T/m)
Here, v is the speed of waves traveling on the string, T is the tension in the string, and m is the linear mass density (mass per unit length) of the string.
What we call the "natural frequency" in here is actually the lowest natural frequency of the string. Given f is the lowest natural frequency of the string, 2f, 3f, 4f, etc. are also natural resonant frequencies of the string, and these are called the "harmonics". This happens because a string has an infinite number of degrees of freedom.
The natural frequency of a string can be found by looking for a sinusoidal solution where the nodes coincide with the fixed ends of the string. This yields the equation:
f = sqrt(T/m) / 2L
Here, T and m are as before, and L is the length of the string. and f is the frequency of the vibration in Hertz (that is what you hear!).
So, we see that the natural frequency depends on three factors: How much tension you apply, whether or not they are Coated guitar strings, how "thick" and how long they are. It seems the original question is incomplete; we can easily assume the same linear mass density, but how about the tension? If we assume the tension is unchanged as well (thus making the question well-defined), the answer is now straightforward. Since length and frequency are clearly inversely proportional, if the length is shortened to one-third its initial length, the frequency will increase to three times its initial value.
Yasar Safkan, Ph.D., Instructor, Yeditepe University, Istanbul, TURKEY
'Knowledge can be communicated, but not wisdom. One can find it, live it, be fortified by it, do wonders through it, but one cannnot communicate and teach it.'