A simple apparatus for demonstrating resonance in an air column is depicted in Fig. 7.59.A vertical pipe open at both ends is partially submerge in water , and a tuning fork vibration at an unknown frequency is placed near the top of the pipe . The length `L` of the air column can be adjusted by moving the pipe vertically . The sound waves generated by the fork are reinforced of the pipe . For a certain pipe , the smallest value of `L` for which a peak occurs in the intensity is `9.00 cm`.
a. What is the frequency of the tuning fork ?
b. What are the values of `L` for the next two resonance conditions?

Explanation

Consider how this problem differs from the preceding problem. In the culvert , the length was fixed and the air column was presented with a mixture of many frequencies . The pipe in this example is presented with one single frequency from the tuning fork , and the lenght of the pipe is varied until resonance is achieved.
This example is a simple substitution problem . Although the pipe is open at its lower end to allow the water to enter , the water's surface acts like a barrier . Therefore , this setup can be modelled as an air column closed at one end .
The fundamental frequency for `L = 0.090 m`:
`f_(1) = (v)/( 4L) = ( 343 m//s )/(( 4( 0.090 m)) = 953 Hz`
Because the tuning fork causes the air column to resonate at this frequency , this frequency must also be that of the tuning fork.
b. Use equation `v = lambda f` to find the wavelength of the sound wave from the tuning fork.
`lambda = (v)/(f) = ( 343 m//s)/( 953 Hz) = 0.360 m`
Notice from Fig. 7.59 (b) that the length of the air column for the second resonance is ` 3 lambda//4`.
`L = 3 lambda//4 = 0.270 m`
Notice from Fig. 7.59(b) that the length of the air column for the third resonance is ` 5 lambda//4`
`L = 5 lambda//4 = 0.450 m`