Answer:
The fundamental frequency is [tex]2f_0[/tex]
Explanation:
For an open-open pipe, the wavelength of the fundamental mode of vibration is equal to twice the length of the pipe:
[tex]\lambda=2L[/tex]
where L is the length of the pipe.
Moreover, from the wave equation we know that the frequency of the fundamental mode is inversely proportional to the wavelength, according to:
[tex]f=\frac{v}{\lambda}[/tex]
where
f is the frequency
v is the speed of the wave
[tex]\lambda[/tex] is the wavelength
Combining the two equations, we get
[tex]f_0=\frac{v}{2L}[/tex]
this means that the fundamental frequency in an open organ pipe is inversely proportional to the length of the pipe.
Here, the pipe is cut in a half, so the new length is
[tex]L'=\frac{L}{2}[/tex]
Therefore, since the speed of the sound wave does not change (it depends only on the medium), the new fundamental frequency will be:
[tex]f'=\frac{v}{2L'}=\frac{v}{2(L/2)}=\frac{v}{L}=2f_0[/tex]
So, the fundamental frequency has doubled.
