A tuning fork with a frequency of 440 Hz is held above a resonance tube that is partially filled with water. Assuming that the speed of sound in air is 342 m/s, for what three smallest heights of the air column will resonance occur? Where will the nodes and anti-nodes occur? (Hint: For resonance to occur, the frequency of the tuning fork must match that of the tube.)
I understand finding the wavelength and finding the height that resonance will occur in the air column but I can't figure out how to find where the nodes and anti nodes occur. Can somebody explain please.
Look at the figures with the closed tube.
http://media.paisley.ac.uk/~davison/labpage/tube/tube.html
To find the three smallest heights of the air column where resonance will occur, we need to calculate the length of the air column that corresponds to the wavelength of the sound produced by the tuning fork.
The formula to calculate the wavelength is:
wavelength = speed of sound / frequency
Given that the frequency of the tuning fork is 440 Hz and the speed of sound in air is 342 m/s, we can substitute these values into the formula:
wavelength = 342 m/s / 440 Hz
wavelength ≈ 0.777 m
The air column in the resonance tube will have nodes and antinodes at certain points. Nodes are points of minimum displacement or no air movement, while antinodes are points of maximum displacement or maximum air movement.
For a tube with one open end (like the resonance tube), the distance from one node to the next node (or from one antinode to the next antinode) is equal to a quarter of the wavelength (λ/4).
To find the three smallest heights of the air column where resonance will occur, we need to find three consecutive nodes. Let's assume the tube is partially filled with water and the height of the water column is "h".
The length of the air column is then equal to the total length of the tube minus the height of the water column:
Length of air column = Total length of tube - height of water column.
For the first node, the air column length will be λ/4. So:
Length of air column for the first node = 0.777 m / 4
For the second node:
Length of air column for the second node = Length of air column for the first node + λ/4
For the third node:
Length of air column for the third node = Length of air column for the second node + λ/4
By plugging in the values and calculating, you can determine the three smallest heights of the air column where resonance will occur.
Regarding the location of the nodes and antinodes:
- The closed end of the tube (where the water is) will always be a node.
- The open end of the tube will always be an antinode.
So, in the resonance tube partially filled with water, the nodes will be located at the closed end (water surface) and at the positions calculated above for the air column heights. The antinodes will be located at the open end of the tube.