Suppose the length of the string is L, then it will vibrates with wave length 2L, 2L/2, 2L/3, 2L/4, ...
The longest wave length (with the lowest frequency f0) is 2L. Its vibration can be described like:
A sin(kx)cos(wt)
with k = 2*Pi/(wave length),
w = 2*Pi*frequency = 2*Pi*f.
Hence w/k = f*(wave length) = wave speed = v.
By just using dimensional analysis, we can find that:
sqrt[(string tension)/(linear mass density)] = sqrt(T/D) has the unit:
sqrt[(kg*m/s*s)/(kg/m)] = m/s, the unit (dimension) of v, so
v is proportional to sqrt[T/D]
Now we see that the frequency is given by:
f = v/(wave Length) ~ sqrt(T/D)/L.
which agrees with our observation of playing Er-Hu.
Let us denote the frequaency of key-note c by f0 (f0 is approximately 131 Hz to 132 Hz), then in the unit of f0, we have the following note-string length-frequency relations;
KEY-NOTE: | DO | RE | Mi | Fa | So | La | Si | Do |
LABEL: | c | d | e | f | g | a | b | c' |
Length in L | 1 | 8/9 | 4/5 | 3/4 | 2/3 | 3/5 | 8/15 | 1/2 |
Freq. in f0 | 1 | 9/8 | 5/4 | 4/3 | 3/2 | 5/3 | 15/8 | 2/1 |