# Why does water 'reflect' a standing sound wave in a tube

I'm reading about (standing) sound waves and one example is that of an vertical pipe that has a standing sound wave in it. It is then filled halfway with water and since the length of the pipe decreases, the required fundamental frequency increases.

Now what I don't understand is how the water just acts like the bottom of the tube. The standing wave should be caused by the reflection of the sound waves that are sent in, but don't the waves continue into the water?

I found a picture that illustrates a standing sound wave on top of the water, where the water surface works like a displacement node.

• If you think this is a problem, why don't you also think the sound waves should travel through the bottom of the tube instead of being reflected, even when there is no water? If you think it is because the tube is somehow "stronger" or "stiffer" than the water, try the experiment with a tube made of paper.... Jan 19 '17 at 23:05
• So with a tube made of paper I would get the same results as with a tube made of concrete? Then why doesn't the sound get reflected by air even? I don't get it. Jan 19 '17 at 23:16
• If you put a sound transducer under water, the sound will get "reflected by the air" at the surface of the water, as @sammygerbil's answer implies. Jan 19 '17 at 23:55

When a wave meets a boundary between 2 media in which the wave speed and/or density is different, there is both transmission and reflection. The bigger the difference in wave speed or density, the higher the % of the wave energy which is reflected.

For normal incidence, the ratio of reflected to incident amplitudes for sound waves in fluids is $R=\frac{\rho_1 c_2 - \rho_2 c_1}{\rho_1 c_2+\rho_2 c_1}$
where $\rho, c$ are the densities and speeds in the two media. See section 2 of these notes from MIT.

The speed of sound in air is about $c_1=330m/s$, while in water it is about $c_2=1484m/s$. The densities are $\rho_1=1.225kg/m^3$ and $\rho_2=1000kg/m^3$. So the reflection coefficient is $R=0.989$ and the % reflection of energy is $R^2=97.8$%. Almost all of the sound energy is reflected by the water surface.

• Great answer, thanks. So basically even if the tube weren't filled with water, a little energy would be lost, and the reflected sound wave would have a displacement amplitude that is just a bit smaller than the original wave's. So there wouldn't actually be a standing wave in the real world, but of course since we use models we can ignore that? Jan 19 '17 at 23:54
• If there is no water in the tube, there would be reflection from the base of the closed tube - also some transmission. If the tube is metal the density and speed of sound are even higher than for water, so the % reflection would be even greater - and the % transmission less. If both ends of the tube are open there can still be some reflection from the open end, but this case is not covered in the notes quoted. ... Standing waves can still form if the amplitudes of incident and reflected waves do not match. Even a small amount of reflection can form a standing wave. Jan 20 '17 at 0:02
• All right thanks. Now just a side question, the displacement amplitude in my picture, is it actually accurate in size or does it just indicate where the displacement is maximum? Jan 20 '17 at 0:16
• The sound amplitude (excess air pressure) varies sinusoidally. It is approx. zero at the air/water boundary and maximum a little above the open end of the tube. Jan 20 '17 at 0:35

Actually most of the sound is reflected from the water surface but not all. Some sound indeed transmits into the water. But for purposes of analyzing, measuring a standing wave it has to do with impedance matching at the air-water boundary. Water has a higher transmissive impedance than air for the sound. Most of the energy is therefore reflected back into and contained within the air space.

And as one of the comments mentioned some is transmitted through the rigid walls.