# Speed of the Standing Wave in A Pipe

I’m studying standing waves and I got stuck with this question. The answer key says ‘A’ is the correct answer, but I think ‘B’ is the correct answer.

My reasoning. The speed of the wave depends on the medium. As we change the frequency of the wave it changes its form to go into the next harmonic so the wavelength changes.

In this question, it says an adjustment is made -say the density of the particles in the environment has changed- so the speed of the standing wave changes. It shouldn’t change its frequency, am I wrong?

The wavelength of standing waves in a pipe would have to depend on the length of the pipe, since we are dividing the length of the pipe between the standing waves. For example, in the third harmonic, we are dividing the length of the pipe, let's say $$L$$, into three parts for each 'node-node' pair in the wave, thus giving us $$\frac{2L}{3}$$ as the wavelength.

This is waves on a string, but the concept is the same.

So, when the density of the particles in the pipe has been changed to change the speed of sound as you said, you may observe that we're still talking about the first harmonic here, which means exactly two antinodes are present on either end of the pipe, and thus the wavelength remains $$2L$$ as the length of the pipe has not changed. Finally, the frequency will increase by the virtue of wavelength remaining constant while the velocity of the medium has increased.

• Therefore, for a given pipe or rope, the wavelength of the $n^{th}$ harmonic will never change, since it completely depends on the length itself. If the velocity of the medium changes, the frequency will alter to adjust to the wavelength, and not the other way round. Commented Apr 15, 2022 at 10:29
The question asks you to consider the first harmonic before and after the adjustment is made. Wavelength $$\lambda$$ will be determined by the geometry of the pipe (length and whether ends are closed or open), which does not change. Speed is related to frequency and wavelength via $$v = f \lambda$$, so frequency $$f$$ must increase by the same factor as the speed of sound $$v$$. Thus A is correct.