"A tuning fork with a frequency of 440 Hz is held above a resonance tube partially filled with water. Assuming that the speed of sound in air is 342 m/s, for what heights of the air column will resonances occur?" f_{1} = 440 Hz v = 342m/s I'm not sure if this is refering to an open pipe, or a closed pipe. For an open pipe, f_{n} = (nv)/(2L) = nf_{1} v/(2L) = f_{1} L = v/(2f_{1}) L = 0.3886 m But this is the length for the sound traveling up and down... i think. So the length is 0.194m. However, the answer gives 3 values, 0.194m, 0.583m, 0.972m. Even for a closed pipe, I there is only 1 value... I must be doing something wrong. I don't think the speed of sound in water is a factor, because my first answer seems to be correct...
This is a closed pipe. There is a node at the bottom and an anti-node at the open end. So resonances occur where the length of the pipe is [itex]L = \lambda/4, 3\lambda/4, 5\lambda/4,... (2n+1)\lambda/4[/itex] Using the universal wave equation: [tex]\lambda = v/f [/tex] Substituting the resonance criterion: [itex]\lambda = 4L/(2n+1) = v/f[/itex] resonance occurs at: [tex]L = (2n+1)v/4f = (2n+1)*342/4*440 = (2n+1).194 m.[/tex] AM
That makes sense, except for why f is kept as the initial frequency. The formula I have is: L = (mv)/(4f_{m}) where m = 1,3,5,..... and f_{m} = mf_{1} thus L = v/(4f_{1}) why do I use only the initial frequency here?
In this case, there is a tuning fork which provides only the fundamental frequency - a pure sine wave. There are no higher frequencies to resonate in the air column. So you are dealing with the fundamental frequency only. AM