How Does Changing Boundary Distance Affect Node Addition in Standing Waves?

[PrinceOfTroy]

Homework Statement

Consider standing waves that are reflecting between two identical boundaries. If the distance between the boundaries is increased by a distance delta L, what value of (delta L / lambda) will add one additional node (and antinode) to the standing wave pattern? (Assume that the wave velocity and frequency are unaffected by the change in distance.)

Homework Equations

L = (n/2)*lambda

The Attempt at a Solution

Isn't it just 1/2, because if we want 2 waves we do 2*(fund_freq) and in this case we want an extra wave?

 

[AvroArrow]

So if we add 1/2 lambda, then the total wavelength would be 1.5 lambda and that would give us 2 waves.

 

[kerimek]

Your solution is correct. When the distance between the boundaries is increased by a distance delta L, the number of nodes in the standing wave pattern will increase by one. This means that the total number of nodes will be (n+1), where n is the number of nodes in the original standing wave pattern. Therefore, the new length L' of the standing wave will be:

L' = ((n+1)/2)*lambda

To find the change in length (delta L), we can subtract the original length L from L':

delta L = L' - L

Substituting L' and L, we get:

delta L = ((n+1)/2)*lambda - (n/2)*lambda

Simplifying this expression, we get:

delta L = (1/2)*lambda

Therefore, the value of (delta L / lambda) that will add one additional node to the standing wave pattern is 1/2. This means that increasing the distance between the boundaries by half of the wavelength will result in an extra node and antinode in the standing wave pattern. This is because the wavelength is directly related to the distance between nodes in a standing wave, and adding half of a wavelength will result in an extra node and antinode.