Question

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

Find two possible values for the length of the pipe.

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(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(v)/(2L)\\L = (nv)/(2f)`

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

`(528Hz)/(660Hz)= (4)/(5)`

`4:5`

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

`4f = 528\\f = (528)/(4)\\f = 132Hz`

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

`L = (nv)/(2f)\\L = ((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 = (nv)/(2f)\\L = ((2)343m/s)/(2(132))\\L = 2.59m`

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