Expert: Scotto - 2/28/2010

Question
The frequency of the note produced by a guitar string is interpreted by
the ear as pitch, and the greater the frequency of the note the higher the pitch. (For
example, the note A, used for tuning instruments, has a frequency of 440 cycles per
second). The frequency f depends on the length of the string L, its linear density
m and its tension T, and is given by the formula:

f= (1/2L)xsqrt((t/m))

(i) How would you expect the frequency to change if L is increased? Check your
answer by calculating the derivative df/dL (assuming that T and m do not
change) and looking at its sign.
(ii) How would you expect the frequency to change if T is increased? Check your
answer by calculating the derivative df/dT (assuming that L and m do not
change) and looking at its sign.
(iii) How would you expect the frequency to change if m is increased? Check your
answer by calculating the derivative df/dm (assuming that T and L do not
change) and looking at its sign.

Answer
In this problem, all of the values are measurements, which means they are all positive.

If the function f= (1/2L)xsqrt((t/m)) is suppose to be f = (1/[2L])x√(t/m) ...
I'll go with that.

(i) The frequency should decrease.
I will take C as the rest of the funciton besides L.
This gives f(L) = C/L.  The derivatve is df/dL = -C/L², which is negative.

(ii) Since t is time, the frequency should increase.
Take the rest of the function outside the squareroot as D.
This give f(t) = D√(t/m).  This gives df/dt = D/(2m√(t/m).
The D is a constant and carries along.  The other part is √(t/m),
so the derivative is [1/(2√(t/m))]d(t/m)/dm, and the last derivative here is 1/m.
This gives df/dt = 2/(2m√(t/m)), which is positive.

(iii) The frequency should decrease with an increase in m.
Let E be the rest of the function outside the squareroot.
Thus, f(m) = E√(t/m), so df/dm = [E/(2√(t/m))]d(t/m)/dm = [E/(2√(t/m))](-t/m²) =
-Et/(2m²√(t/m))].  The derivative is negative, so the function is decreasing.