An organ pipe, open at both ends, resonates at its first resonant length with a frequency of 128 Hz. What is the length of the pipe if the speed of sound is 346m/s?
pipe is half a wave long? (must be, antinode at both ends.)
what is period of this wave?
T = 1/f = 1/128 second
goes a wavelength in 1/128 second
346 m/s (1/128) = wavelength
so wavelength = 346/128 meters
so
1/2 wavelength = 346 /256 meters
s
Do not hurry me. I get very nervous and upset.
Well, if the organ pipe is open at both ends, I guess you could say it's living life on the edge, not wanting to commit to one opening. As for its first resonant length, we need to find the length that creates a frequency of 128 Hz. Time to do some math!
The formula to calculate the frequency of a resonant pipe is f = (v / 2L), where f is the frequency, v is the speed of sound, and L is the length of the pipe.
Plugging in the given values, we have:
128 Hz = (346 m/s) / (2L)
Now, let's solve for L.
2L = (346 m/s) / (128 Hz)
L = (346 m/s) / (128 Hz) / 2
Calculating this out, we get L = 1.35156 m.
So, the length of the organ pipe is approximately 1.35156 meters. Keep honking those tunes, Mr. Organ Pipe!
To find the length of the organ pipe, we can use the formula for the resonant frequency of an open pipe:
\[ f = \dfrac{v}{2L} \]
where:
- \(f\) is the frequency of the resonance,
- \(v\) is the speed of sound, and
- \(L\) is the length of the pipe.
We are given that the frequency (\(f\)) is 128 Hz and the speed of sound (\(v\)) is 346 m/s. We need to solve for the length of the pipe (\(L\)).
Rearranging the formula, we can solve for \(L\):
\[ L = \dfrac{v}{2f} \]
Plugging in the given values:
\[ L = \dfrac{346}{2 \times 128} \]
Calculating the result:
\[ L = \dfrac{346}{256} \]
\[ L \approx 1.35 \, \text{meters} \]
Therefore, the length of the organ pipe is approximately 1.35 meters.
ASAP
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