Why Do Standing Waves Require Specific Wavelength Conditions?

[suspenc3]

Why do standing waves only exist when $$\lambda=2L/n$$? For example why don't they exist when $$\lambda=4L$$

Not really sure but if $$\lambda$$ equals 4L then there is only going to be one "loop", I am guessing this has somehting to do with it but I am not sure how to explain it.

 

[chroot]

Generally, a standing wave exists when the reflection of the wave from one end of the cavity reinforces the wave already present in the cavity. The generally means the ends of the cavity need to be nodes to support a standing wave. When the wavelength is one-quarter of the length of the cavity, it will not be self-reinforcing, and will decay. This is because nothing "special" happens at one-quarter of the wavelength; compare to the point at half the wavelength, where there is a node.

- Warren

 

[m2003]

Standing waves only exist when the wavelength (\lambda) is equal to 2 times the length of the medium (L) divided by the number of nodes (n). This is because standing waves are formed by the interference of two waves traveling in opposite directions. When the wavelength is equal to 2L/n, the two waves will interfere constructively at the nodes (points of no displacement) and destructively at the antinodes (points of maximum displacement), creating a stable pattern of standing waves.

When the wavelength is not equal to 2L/n, the two waves will not interfere in a consistent manner, resulting in a constantly changing pattern that does not meet the criteria for a standing wave. This is why standing waves do not exist when \lambda=4L, as there will only be one complete loop instead of the required two loops for a standing wave.

In other words, the wavelength must be precisely related to the length of the medium and the number of nodes in order for the constructive and destructive interference to occur at the same points and create a stable standing wave pattern.