Two organ pipes, open at one end but closed at the other, are each. 1.14 m long. One is now lengthened by 2.00 cm. Find the frequency of the beat they produce when playing together in their fundamental.
f1=(344m/s) / (4*1.14m)=75.4386Hz
f2=(344m/s) / (4*(1.14m+.02m))=74.1379Hz
f=f1-f2=1.3007Hz
Going to be about 1-2 hertz fyi, 345/pipe1 345/pipe2 and difference
Well, it seems like these organ pipes are having a bit of a length competition. One's feeling a little insecure and decided to elongate itself by 2.00 cm. Now let's see what happens when these two pipes try to harmonize.
To find the frequency of the beat they produce, we need to determine the difference in their frequencies. But before we go any further, we need to find the original frequencies of each pipe.
Since the pipes are open at one end and closed at the other, they both produce the odd harmonics only. The fundamental frequency (1st harmonic) for an open-closed pipe is given by the formula:
v/4L
where v is the speed of sound and L is the length of the pipe.
Let's calculate the original frequency of each pipe:
Pipe 1:
Frequency = v / (4 * 1.14)
Pipe 2:
Frequency = v / (4 * 1.14)
Now that we have the original frequencies, let's calculate the difference in their frequencies:
Δf = |Frequency of Pipe 1 - Frequency of Pipe 2|
And finally, the frequency of the beat they produce is twice the difference in frequencies:
Frequency of the beat = 2 * Δf
Now, you may be wondering, why all this complicated math? Well, when the pipes are playing together, they create a rhythm that sounds like a beat. It's like having two clowns trying to play the same tune but with slightly different timing - and that's what we call a beat!
So there you have it, the frequency of the beat these organ pipes will produce when playing together in their fundamentals. Just remember, they may be pipes, but they're not chill enough to play in perfect harmony.
To find the frequency of the beat produced when two organ pipes play together in their fundamental, we need to calculate the difference in frequencies between the two pipes.
The frequency of a pipe is inversely proportional to its length.
Let's first find the initial frequency of the two pipes. Since both pipes are 1.14 m long, they have the same length, which means they will have the same initial frequency.
To find the initial frequency, we need to know the speed of sound in air. Let's assume it is approximately 343 meters per second.
Using the formula for the frequency of an open-open pipe:
Frequency = (speed of sound) / (2 * length)
Frequency = 343 m/s / (2 * 1.14 m)
Frequency = 150.88 Hz (approximately)
Now let's calculate the frequency of the lengthened pipe. The length of this pipe is 1.14 m + 2.00 cm, which we need to convert to meters.
2.00 cm = 0.02 m
Length of lengthened pipe = 1.14 m + 0.02 m = 1.16 m
Using the same formula as before:
Frequency of lengthened pipe = 343 m/s / (2 * 1.16 m)
Frequency of lengthened pipe = 147.84 Hz (approximately)
To calculate the beat frequency, we need to find the difference between the frequencies of the two pipes:
Beat frequency = |Frequency of lengthened pipe - Initial frequency|
Beat frequency = |147.84 Hz - 150.88 Hz|
Beat frequency = 3.04 Hz (approximately)
Therefore, the frequency of the beat produced when the two organ pipes play together in their fundamentals is 3.04 Hz.