The fundamental of an organ pipe that is closed at one end and open at the other end is 261.6 Hz ( middle C ). The second harmonic of an organ pipe that is open at both ends has the same frequency. What are the lengths of the pipe?
I tried solving it and I got 0.65 and 0.98 and they are wrong .. how can I solve it :S
What speed of sound are you assuming?
To solve this problem, we need to use the formula for the fundamental frequency of a closed-closed or open-open organ pipe:
f = (nv) / (2L)
where:
- f is the frequency of the harmonic,
- n is the harmonic number,
- v is the speed of sound, and
- L is the length of the pipe.
In this case, we are given that the fundamental frequency (n=1) of the closed-open pipe is 261.6 Hz. We are also told that the second harmonic (n=2) of the open-open pipe is the same frequency.
First, let's find the speed of sound. The speed of sound in air is approximately 343 meters per second.
Now, let's solve for the length of the closed-open pipe:
261.6 = (1*343) / (2L1) [Substitute the given values]
523.2L1 = 343 [Cross multiply]
L1 = 343 / 523.2 [Divide both sides by 523.2]
L1 ≈ 0.655 meters
Now, let's solve for the length of the open-open pipe:
261.6 = (2*343) / (2L2) [Substitute the given values]
523.2L2 = 686 [Cross multiply]
L2 = 686 / 523.2 [Divide both sides by 523.2]
L2 ≈ 1.31 meters
Therefore, the lengths of the closed-open and open-open pipes are approximately 0.655 meters and 1.31 meters, respectively.
To solve this problem, we can use the formula for the frequency of a vibrating string or organ pipe:
f = (nv)/(2L),
where f is the frequency, n is the harmonic number, v is the speed of sound, and L is the length of the pipe.
Given:
- Fundamental frequency (first harmonic) of closed-open pipe (n = 1): f1 = 261.6 Hz
- Second harmonic of open-open pipe (n = 2): f2 = 261.6 Hz
We know that for a closed-open pipe, the fundamental frequency is the first harmonic, while for an open-open pipe, the fundamental frequency is the second harmonic. Therefore, we have:
f1 = (1v)/(2L1),
f2 = (2v)/(2L2).
Since the second harmonic of the open-open pipe has the same frequency as the fundamental frequency of the closed-open pipe, we can equate the two expressions:
f1 = f2,
(1v)/(2L1) = (2v)/(2L2).
We can now solve for the lengths L1 and L2.
1. Start by cancelling out the factors of v and 2:
1/(2L1) = 2/(2L2).
2. Cross-multiply:
2L1 = 4L2.
3. Divide both sides of the equation by 2:
L1 = 2L2.
Now, we can substitute this relationship back into one of the original equations to solve for the lengths.
Let's solve for L1:
f1 = (1v)/(2L1).
Substitute the known value for the frequency f1:
261.6 Hz = (1v)/(2L1).
We can simplify this equation further:
L1 = (1v)/(261.6 Hz * 2).
Now, you can calculate the length of the closed-open pipe (L1) using this equation.