Question

A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).

Answer 1

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

$f = n\frac{v}{2L}$

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,

$f = n\frac{v}{2L}\\L = \frac{nv}{2f}$

The radius between the two frequencies would be 4 to 5,

$\frac{528Hz}{660Hz}= \frac{4}{5}$

$4:5$

Therefore the frequencies are in the ratio of natural numbers.  That is

$4f = 528\\f = \frac{528}{4}\\f = 132Hz$

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

$L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m$

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

$L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m$

Therefore the length of the pipe in the second harmonic is 2.6m