If a string vibrates at the fundamental frequency of 528 Hz and also produces an overtime with a frequency of 1,056 Hz this over
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For a pipe closed at one end, the second overtone is the 3rd natural frequency as shown in the diagram.
For the fundamental frequency,
λ = 4L
where
λ = wavelength
L = the length of the tube
For the second overtone (3rd frequency)
λ = (4/3)L
Because L = 6.8 m, therefore
λ = (4/3)*(6.8 m) = 9.0667 m
The distance between a node and an adjacent antinode is λ/2, which is
9.0667/2 = 4.533 m
Answer: 4.5 m
For the fundamental frequency,
λ = 4L
where
λ = wavelength
L = the length of the tube
For the second overtone (3rd frequency)
λ = (4/3)L
Because L = 6.8 m, therefore
λ = (4/3)*(6.8 m) = 9.0667 m
The distance between a node and an adjacent antinode is λ/2, which is
9.0667/2 = 4.533 m
Answer: 4.5 m