Find Length of Closed Pipe for Equal Frequency

[diracdelta]

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, f_n=n*f₁, f₁=v/(2L)
Closed pipe; f_n= f*f₁, f₁ = v/(4L)
f_beat = 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

f_beat = 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?

 

[NascentOxygen]

Now we need to find the length of open pipe.

I think you mean closed pipe? Better check that calculation, too, using 555Hz. ☹

 

[ehild]

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

f_beat = 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?

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 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.

 

[diracdelta]

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 :)