Find Length of Closed Pipe for Equal Frequency

In summary, the conversation discusses finding the length of a closed pipe in order to have both an open and closed pipe produce equal frequencies. The equation for the open pipe is fn=n*f1 and the equation for the closed pipe is fn=f*f1, with fbeat=f2-f1. The calculation for the length of the open pipe was incorrect due to rounding errors, but was corrected to be 0.1486 m. The final calculation for the length of the closed pipe was also incorrect due to rounding errors, but was corrected to be 0.15 m.
  • #1
diracdelta
55
0

Homework Statement


Open and closed pipe, give 5 beats per second.
Open pipe is 30 cm long and gives a tone of frequency f0.
Speed of sound is 330 m/s.
How long do we need to extend closed pipe so both pipes give equal frequencies?

Homework Equations


Open pipe, fn=n*f1, f1=v/(2L)
Closed pipe; fn= f*f1, f1 = v/(4L)
fbeat = f2 - f1

The Attempt at a Solution


For open pipe, I assume that f0 is f1, or basic frequency.
f=v/2L=330 ms-1 / 0,6 m = 550 Hz

fbeat = f2 - f1
5 = f2 - 550 Hz
f2 = 555 Hz.

Now we need to find the length of open pipe.
4L = v/f2
l= v/4f2= 0,15 m

Since i have the length of closed pipe, motive is to adjust it so we have frequency of 550 Hz.
4L = v/f2,
4*L = 330 ms-1/ 550 Hz
4*L = 0,6 m
L = 0,15

What did I do wrong?
 
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  • #2
Now we need to find the length of open pipe.
I think you mean closed pipe? Better check that calculation, too, using 555Hz.
 
  • #3
diracdelta said:

The Attempt at a Solution


For open pipe, I assume that f0 is f1, or basic frequency.
f=v/2L=330 ms-1 / 0,6 m = 550 Hz

fbeat = f2 - f1
5 = f2 - 550 Hz
f2 = 555 Hz.

Now we need to find the length of open pipe.

You ment closed pipe, didn't you?

diracdelta said:
4L = v/f2
l= v/4f2=0,15 m
l= v/(4f2)=330/(4*555)=0.1486 m
You rounded off too early, too much.
diracdelta said:
Since i have the length of closed pipe, motive is to adjust it so we have frequency of 550 Hz.
4L = v/f2,
4*L = 330 ms-1/ 550 Hz
4*L = 0,6 m
L = 0,15

What did I do wrong?

You rounded off too much.
 
  • #4
NascentOxygen said:
I think you mean closed pipe? Better check that calculation, too, using 555Hz.
Yes. lapsus linguae :D

ehild said:
You ment closed pipe, didn't you?l= v/(4f2)=330/(4*555)=0.1486 m
You rounded off too early, too much.

You rounded off too much.

:(
I did it again, i got same result like yours.

Thanks for help guys.
You are the best :)
 
  • #5


Your calculations are correct, but the length of the closed pipe should be 0.3 m, not 0.15 m. This is because the length of the open pipe is 0.3 m, so the total length of the two pipes should be 0.3 m + 0.3 m = 0.6 m. Therefore, the length of the closed pipe should be 0.6 m - 0.3 m = 0.3 m.
 

1. How do you calculate the length of a closed pipe for equal frequency?

The length of a closed pipe for equal frequency can be calculated using the formula: L = (n/2) * v/f, where L is the length of the pipe, n is the number of nodes created by the standing wave, v is the speed of sound, and f is the frequency of the sound wave.

2. What is the significance of equal frequency in a closed pipe?

Equal frequency in a closed pipe refers to the standing wave pattern that is created when the frequency of the sound wave matches the natural frequency of the pipe. This results in resonance, where the amplitude of the sound wave is amplified, allowing for better sound quality.

3. What factors affect the length of a closed pipe for equal frequency?

The length of a closed pipe for equal frequency is affected by the speed of sound, the frequency of the sound wave, and the number of nodes created by the standing wave. Additionally, the material and thickness of the pipe can also impact the length.

4. How does the number of nodes affect the length of a closed pipe for equal frequency?

The number of nodes created by the standing wave is directly proportional to the length of the pipe. This means that the longer the pipe, the more nodes that will be created and vice versa. This relationship is important in determining the length of a closed pipe for equal frequency.

5. Can the length of a closed pipe be adjusted for different frequencies?

Yes, the length of a closed pipe can be adjusted to produce different frequencies by changing the number of nodes created by the standing wave. This can be achieved by either changing the length of the pipe or adjusting other factors, such as the speed of sound or the material of the pipe.

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