# How can frequency of acoustic wave not change?

I am already familiar with the notion of extra waves coming out of nowhere or waves piling up, but consider this- A sound wave passes from a less dense to a denser medium(so speeds up) . Assuming frequency is same, the wavelength increases. Since time period is inversely related to wavelength, so it decreases. Since f=1/t, f changes. What is happening?

• The effect of density is to decrease the speed of sound. But you cannot tell if it speeds up or slows down just from density. The elastic properties of the medium matter too.
– nasu
Nov 12, 2021 at 12:01
• i am asking that is speed increase, wavelength increase, so time period decrease, so frequency increase... HOW Nov 12, 2021 at 12:14
• The period is not inversely related to wavelength. You got this wrong too.
– nasu
Nov 12, 2021 at 12:18

The frequency (not the period) is inversely related to the wavelength, so you can write $$f = \frac{1}{T} \propto\frac{1}{\lambda},$$ where $$\lambda$$ is the wavelength, $$T$$ is the period, and $$f$$ is the frequency. However, proportionality is not equality. There may be other factors involved. In fact, equality may be written as $$f = \frac{1}{T} = \frac{c}{\lambda},$$ where $$c$$ is the wave speed. If the frequency is held constant, which implies that the period is held constant, then changing the wave speed changes the wavelength. You can actually use this relationship to figure out how the wavelength changes when you change wave speeds.
Incidentally, the relationship $$\lambda f=c$$ is true for all linear waves, where the mathematical concept of linearity is roughly equivalent to the physical concept of small amplitudes. Thus, the above explanation works for water waves and light, as well.
• The period $T$ and the frequency $f$ are, by definition, the inverse of each other, without any additional constants. If you hold $f$ or $T$ constant and change the wave speed your wavelength will change in proportion. I updated the answer to emphasize that $T$ and $f$ give the same information. Nov 12, 2021 at 12:31
• No, because the wavelength goes down at the same rate as the wave speed, so the ratio $c/\lambda$ does not change. Nov 12, 2021 at 12:34