To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.
By definition the oscillation frequency is defined as
[tex]f = n\frac{v}{2L}[/tex]
Where
v = speed of sound
L = Length of the pipe
n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)
Re-arrange to find L,
[tex]f = n\frac{v}{2L}\\L = \frac{nv}{2f}[/tex]
The radius between the two frequencies would be 4 to 5,
[tex]\frac{528Hz}{660Hz}= \frac{4}{5}[/tex]
[tex]4:5[/tex]
Therefore the frequencies are in the ratio of natural numbers. That is
[tex]4f = 528\\f = \frac{528}{4}\\f = 132Hz[/tex]
Here f represents the fundamental frequency.
Now using the expression to calculate the Length we have
[tex]L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m[/tex]
Therefore the length of the pipe is 1.3m
For the second harmonic n=2, then
[tex]L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m[/tex]
Therefore the length of the pipe in the second harmonic is 2.6m
