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Shortening a guitar string to one-third its initial length will change its natural frequency by what factor?
Asked by: elva b.
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.