Answer: 0.318 Hz.
Explanation: (A). In an open organ pipe, the frequencies of the harmonics are given by the formula:
f = (nv) / (2L)
Where:
f is the frequency of the harmonic,
n is the harmonic number,
v is the speed of sound in air, which is approximately 343 m/s at room temperature,
and L is the length of the pipe in meters.
Converting the length from centimeters to meters, we have:
L = 78.6 cm = 78.6 / 100 = 0.786 m
To find the fourth harmonic frequency (n = 4) of the open organ pipe, we plug in the values into the formula:
f = (4 * 343) / (2 * 0.786)
Calculating the above expression, we find:
f ≈ 869.18 Hz
Therefore, the fourth harmonic frequency of the open organ pipe is approximately 869.18 Hz.
(B). For a closed organ pipe, the frequencies of the harmonics are given by the formula:
f = (nv) / (4L)
Using the given values:
L = 68.5 cm = 68.5 / 100 = 0.685 m
n = 3 (the first overtone or second harmonic)
Substituting the values into the formula, we have:
f = (3 * 343) / (4 * 0.685)
Computing the above expression, we find:
f ≈ 500.00 Hz
Therefore, the first overtone (or n = 3 harmonic) frequency of the closed organ pipe is approximately 500.00 Hz.
(C). For an open organ pipe resonating at a specific frequency, the fundamental frequency is given by the formula:
f = (v) / (2L)
Using the given resonant frequency of 539 Hz, we can solve for the fundamental frequency by rearranging the formula as follows:
2L = v / f
L = (v / f) / 2
Substituting the known values:
v = 343 m/s
f = 539 Hz
Calculating the above expression, we find:
L = (343 / 539) / 2
L ≈ 0.318 m
Therefore, the fundamental frequency of the organ pipe with the second-shortest length (open at both ends) is approximately 0.318 Hz.