Open and closed pipe

  • Thread starter diracdelta
  • Start date
  • #1
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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 lenght of open pipe.
4L = v/f2
l= v/4f2= 0,15 m

Since i have the lenght 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?
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
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Now we need to find the lenght of open pipe.
I think you mean closed pipe? Better check that calculation, too, using 555Hz. ☹
 
  • #3
ehild
Homework Helper
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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 lenght of open pipe.
You ment closed pipe, didn't you?

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.


Since i have the lenght 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
55
0
I think you mean closed pipe? Better check that calculation, too, using 555Hz. ☹
Yes. lapsus linguae :D

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 :)
 

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