Calculate the length of a closed organ pipe with a 680 Hz fundamental frequency. What is the wavelength of the 4th overtone?
Half a wave fits in the pipe at fundamental, amplitude is zero at both ends.
I do not know what you are using for the speed of sound. Call it s meters/second
period of wave T= 1/680 seconds
wave goes distance 2 L in 1/680 seconds
so
2 L = s * T = s/680
L = s /1360
first harmonic, one whole wave fits in pipe
second harmonic, 1.5 waves fit in pipe
third harmonic 2 waves fit in pipe
fourth harmonic 2.5 waves fit in pipe
so wavelength = that 2 L we had /2.5
sorry I have it closed at both ends
Usually open at one, closed at the other
fundamental 1/4 wave fits in pipe
4 L = s * T = s/680 = wavelength of fundamental
L = s /2720
first harmonic, 3/4 wave fits in pipe
second harmonic, 1 1/4 waves fit in pipe
third harmonic 1 3/4 waves fit in pipe
fourth harmonic 2 1/4 waves fit in pipe
so wavelength = original wavelength we had /2.25
To calculate the length of a closed organ pipe with a given fundamental frequency, we can use the formula:
λ = 4L / n
Where:
λ = Wavelength
L = Length of the pipe
n = Harmonic number
Given that the fundamental frequency (1st harmonic) is 680 Hz, we can find the length of the pipe using the above formula for the 1st harmonic:
680 = 4L / 1
Rearranging the equation to solve for L:
L = 680 / 4
L = 170
Therefore, the length of the closed organ pipe is 170 units of length (e.g., centimeters, meters, etc.).
To find the wavelength of the 4th overtone (5th harmonic), we can use the same formula:
λ = 4L / n
Substituting the values:
λ = 4 * 170 / 5
λ = 680 / 5
λ = 136
Therefore, the wavelength of the 4th overtone is 136 units of length (e.g., centimeters, meters, etc.).
To calculate the length of a closed organ pipe, we need to use the formula:
Length = (2n - 1) * (v/4f)
Where:
n is the harmonic number (1 for the fundamental frequency, 2 for the 1st overtone, 3 for the 2nd overtone, and so on),
v is the speed of sound in air (approximately 343 m/s at room temperature),
and f is the frequency of the harmonic.
Let's start by finding the length of the closed organ pipe for the fundamental frequency (1st harmonic):
Length = (2 * 1 - 1) * (343 / 4 * 680)
= (1) * (343 / 2720)
= 0.126 meters (rounded to three decimal places)
Now, let's find the wavelength of the 4th overtone. The wavelength (λ) can be calculated using the following formula:
λ = 2L / n
Where L is the length of the closed organ pipe and n is the harmonic number.
For the 4th overtone (n = 4):
λ = 2 * 0.126 / 4
= 0.126 / 2
= 0.063 meters (rounded to three decimal places)
Therefore, the wavelength of the 4th overtone for a closed organ pipe with a fundamental frequency of 680 Hz is approximately 0.063 meters.