An organ pipe that is open a t both ends has a fundamental frequency of 382 Hz. What is the length of an closed end pipe that has a fundamental frequency exactly one octave higher? Report your answer in m.
new freq: 2*382hz
for one end closed..
L/4=lambda=speedsound/freq
solve for L
To solve this problem, we need to use the formula for the fundamental frequency of an open cylindrical pipe:
f = v / (2L)
where f is the frequency, v is the speed of sound in air, and L is the length of the pipe.
Given that the fundamental frequency of the open pipe is 382 Hz, we can rearrange the formula to solve for L:
L = v / (2f)
Now, let's consider the closed end pipe. The fundamental frequency of a closed pipe is one octave higher, which means it is twice the frequency of the open pipe:
f_closed = 2 * 382 Hz = 764 Hz
Using the same formula, we can solve for the length of the closed pipe:
L_closed = v / (2f_closed)
Since we are comparing the two lengths, we can set up a ratio:
L_closed / L = (v / (2f_closed)) / (v / (2f))
The v and (2/2) terms cancel out, simplifying the equation to:
L_closed / L = f / f_closed
Plugging in the given values:
L_closed / L = 382 Hz / 764 Hz
L_closed / L = 1/2
Now, we can solve for L_closed by multiplying both sides of the equation by L:
L_closed = (1/2) * L
Therefore, the length of the closed end pipe is half the length of the open pipe.
Since we don't have the length of the open pipe (L), we cannot determine the exact length of the closed end pipe. However, we know that it will be half the length of the open pipe.
To find the length of the closed end pipe that has a fundamental frequency exactly one octave higher than the open-end pipe, we need to use the concept of harmonics.
The fundamental frequency is the lowest frequency produced by the pipe, and one octave higher means that the frequency is doubled. Therefore, the fundamental frequency of the closed end pipe would be 382 Hz × 2 = 764 Hz.
In a closed end pipe, the fundamental frequency (f1) is related to the length (L) by the equation:
f1 = (v / 2L)
Where:
f1 = fundamental frequency
v = speed of sound
To solve for L, we need to rearrange the equation:
L = v / (2f1)
Now we can substitute the values:
v ≈ 343 m/s (the approximate speed of sound in air at room temperature)
f1 = 764 Hz
L = 343 m/s / (2 × 764 Hz)
L ≈ 0.2246 meters
Therefore, the length of the closed end pipe that has a fundamental frequency exactly one octave higher is approximately 0.2246 meters.