When a metal pipe is cut into two pieces, the lowest resonance frequency in one piece is 244Hz and that for the other is 454Hz. What resonant frequency would have been produced by the original length of pipe? (Assume the speed of sound in air is 345 m/s)
b). How long was the original pipe?
To find the resonant frequency of the original length of the pipe, we need to determine the lengths of the two pieces first.
Let's denote the length of the first piece as L1 and the length of the second piece as L2.
We know that the lowest resonance frequency in one piece is 244Hz, so we can use the formula for the resonant frequencies of closed-closed pipe resonators:
f1 = (n1 * v) / (4 * L1)
where f1 is the frequency, n1 is the harmonic number (in this case, 1 for the lowest resonance), v is the speed of sound in air, and L1 is the length of the first piece.
Similarly, for the second piece:
f2 = (n2 * v) / (4 * L2)
where f2 is the frequency, n2 is the harmonic number (again, 1 for the lowest resonance), v is the speed of sound in air, and L2 is the length of the second piece.
Since the two pieces were originally part of the same pipe, the frequency and speed of sound are the same for both pieces:
f1 = f2
n1 * v / (4 * L1) = n2 * v / (4 * L2)
L1 / L2 = n1 / n2
From the given information, we have f1 = 244Hz, f2 = 454Hz, n1 = 1, and n2 = 1.
Plugging these values into the equation, we can solve for L1/L2:
L1 / L2 = 1 / 1
L1 / L2 = 1
Since the lengths of the two pieces are in a ratio of 1:1, we can conclude that they are equal:
L1 = L2
Now, to find the resonant frequency of the original length of the pipe, we can use either L1 or L2 since they are the same length.
Let's use L1 = L2 = L for simplicity.
We can rearrange the formula for f1 to solve for L:
4 * L = (n1 * v) / f1
L = (n1 * v) / (4 * f1)
Plugging in the values we have, n1 = 1, v = 345 m/s, and f1 = 244Hz:
L = (1 * 345) / (4 * 244)
L ≈ 0.354 m
Therefore, the resonant frequency of the original length of the pipe would have been 244Hz, and the length of the original pipe is approximately 0.354 meters.