Given:
The length of the pipe is L.
The third harmonic has teh wavelength,
[tex]\frac{2L}{3}[/tex]
To find:
The type of the pipe
Explanation:
The wavelength of the harmonics for a pipe closed at one end is,
[tex]\lambda=\frac{4L}{2n-1}[/tex]
The wavelength of the harmonics for a pipe open at both the ends is,
[tex]\lambda=\frac{2L}{n}[/tex]
Here,
[tex]n=1,\text{ 2, ......}[/tex]
The given wavelength is,
[tex]\frac{2L}{3}[/tex]
which resembles the form,
[tex]\frac{2L}{n}[/tex]
Hence, the pipe is open at the other end.
