Animated Solution for Physics - Waves: A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is dipped vertically in water, so that half of it is in water. The fundamental frequency of the air column is now

Sometimes in physics, changing the physical setup of a system leads to a surprisingly elegant and unchanged result. This classic problem of an organ pipe dipped in water is a perfect example of how boundary conditions dictate the behavior of standing waves.

Imagine an open organ pipe of length $l$. When sound waves travel through this pipe, they reflect off the open ends. Because the ends are open to the atmosphere, the air molecules are free to oscillate maximally, creating displacement antinodes at both ends. Exactly in the middle of the pipe, a displacement node forms where the air molecules are stationary.

The fundamental frequency $f$ of an open pipe is determined by the condition that the length of the pipe $l$ must accommodate exactly half a wavelength ($\frac{\lambda }{2}=l$). Therefore, the wavelength is $\lambda =2l$. Using the wave equation $v=f\lambda$, we get the fundamental frequency:

Now, we introduce a twist. We dip this pipe vertically into water until exactly half of it is submerged. What happens to the air column inside?

The water surface acts as a rigid, impenetrable boundary for the sound waves. Air molecules cannot push the dense water out of the way, so their longitudinal displacement at the water surface is forced to be zero. This creates a displacement node at the water level. The top end of the pipe remains open, so it still forms an antinode.

Physically, our open pipe has transformed into a closed organ pipe. Furthermore, because half of the pipe is underwater, the effective length of the vibrating air column is now $l^{\prime }=\frac{l}{2}$.

For a closed organ pipe, the fundamental mode consists of one node at the closed end and one antinode at the open end. This corresponds to a quarter of a wavelength fitting inside the pipe ($\frac{{\lambda }^{\prime }}{4}=l^{\prime }$). The fundamental frequency $f^{\prime }$ for a closed pipe is given by:

Let's substitute our new effective length $l^{\prime }=\frac{l}{2}$ into the closed pipe formula:

Notice how the $4$ in the numerator and the $2$ in the denominator interact. They simplify beautifully:

Look closely at this result. It is exactly the same expression we derived for the original open pipe! Therefore, we can conclude that:

Despite cutting the length of the air column in half and changing one of the boundaries from open to closed, the fundamental frequency remains completely unchanged. The halving of the length perfectly compensates for the shift from a half-wavelength resonance to a quarter-wavelength resonance.