What are the first, second, and third frequencies audible from a 20 cm long organ pipe when (A) only one end is open and when (B) both ends are open. The speed of sound through the air inside the organ pipe is 343m/s.
To determine the frequencies of the first, second, and third harmonics for an organ pipe, we can use the formula:
𝑓(𝑛) = (𝑛𝑣) / (2𝐿)
Where:
𝑓(𝑛) is the frequency of the nth harmonic,
𝑛 is the harmonic number,
𝑣 is the speed of sound, and
𝐿 is the length of the organ pipe.
(A) When only one end of the organ pipe is open:
In this case, the open end of the pipe acts as an antinode, meaning that there is an alternating pattern of high and low pressure at this point. The closed end acts as a node, where there is always low pressure. Therefore, the length of the pipe corresponds to half of a wavelength.
The first harmonic will have a single antinode and a wavelength twice the length of the pipe, so:
𝜆(1) = 2𝐿
𝑓(1) = (𝑛𝑣) / (2𝐿) = (1 * 𝑣) / (2𝐿)
The second harmonic will have two antinodes and a wavelength equal to the length of the pipe, so:
𝜆(2) = 𝐿
𝑓(2) = (2 * 𝑣) / (2𝐿) = 𝑣 / 𝐿
The third harmonic will have three antinodes and a wavelength two-thirds the length of the pipe, so:
𝜆(3) = (2/3)𝐿
𝑓(3) = (3 * 𝑣) / (2𝐿)
(B) When both ends of the organ pipe are open:
In this case, both ends of the pipe act as antinodes. The length of the pipe corresponds to one full wavelength.
The first harmonic will have a single antinode and a wavelength equal to the length of the pipe, so:
𝜆(1) = 𝐿
𝑓(1) = (𝑛𝑣) / (2𝐿) = (1 * 𝑣) / (2𝐿)
The second harmonic will have two antinodes and a wavelength half the length of the pipe, so:
𝜆(2) = (1/2)𝐿
𝑓(2) = (2 * 𝑣) / (2𝐿) = 𝑣 / 𝐿
The third harmonic will have three antinodes and a wavelength one-third the length of the pipe, so:
𝜆(3) = (1/3)𝐿
𝑓(3) = (3 * 𝑣) / (2𝐿)
Now, substituting 𝑣 = 343m/s and 𝐿 = 20cm = 0.2m into the equations above, we can calculate the first, second, and third frequencies for both cases.
To find the first, second, and third frequencies audible from a 20 cm long organ pipe, we can use the formula:
fn = n * v / 2L
Where:
- fn is the frequency of the nth harmonic
- n is the harmonic number (1 for fundamental, 2 for second harmonic, 3 for third harmonic, and so on)
- v is the speed of sound in the medium (in this case, the speed of sound through the air inside the organ pipe)
- L is the length of the organ pipe
Given:
- Length of the organ pipe (L) = 20 cm = 0.2 m
- Speed of sound through the air (v) = 343 m/s
(A) When only one end of the organ pipe is open (closed at the other end):
In this case, we have a quarter-wavelength resonator. Only odd harmonics are present.
For the first frequency (fundamental), n = 1:
f1 = 1 * v / 4L
For the second frequency (second harmonic), n = 3:
f2 = 3 * v / 4L
For the third frequency (third harmonic), n = 5:
f3 = 5 * v / 4L
(B) When both ends of the organ pipe are open:
In this case, we have a half-wavelength resonator. All harmonics are present.
For the first frequency (fundamental), n = 1:
f1 = 1 * v / 2L
For the second frequency (second harmonic), n = 2:
f2 = 2 * v / 2L
For the third frequency (third harmonic), n = 3:
f3 = 3 * v / 2L
Now, we can substitute the given values into the formulas above to calculate the frequencies.
(A) When only one end is open:
f1 = 1 * 343 / (4 * 0.2)
f2 = 3 * 343 / (4 * 0.2)
f3 = 5 * 343 / (4 * 0.2)
(B) When both ends are open:
f1 = 1 * 343 / (2 * 0.2)
f2 = 2 * 343 / (2 * 0.2)
f3 = 3 * 343 / (2 * 0.2)
Now, you can substitute the values into the equations and calculate the frequencies for both cases.