Question

Suppose the frequency of a note on an organ is 18 Hz. What is the shortest organ pipe with both ends open that will resonate at this frequency

Answer 1

9.53 m

Explanation:

The computation of shortest organ pipe with both ends open that will resonate at this frequency is shown below:-

`\lambda = (velocity)/(frequency)`

`= (343)/(18)`

= 19.06 m

Now the

Shortest organ pipe with both ends open is

=
`(\lambda)/(2)`

`= (19.06)/(2)`

= 9.53 m

Basically we applied the above formulas so that first we easily determined the shortest organ pipe for both ends at this frequency