# Beat frequency and length question. (grade 11)

1. May 23, 2015

### Ethan_Tab

1. The problem statement, all variables and given/known data
Two pipes, Identical in length and closed at one end are producing notes. You are very, very annoyed because the notes are creating an audible beat frequency of 20Hz. If one pipe is producing a note at 10Hz and the temperature is 15C (which correlates to a speed of sound of 340.39m/s), what is the length of the two pipes?

2. Relevant equations

Beat frequency= abs(ƒ_2-ƒ_1)
L(n)=[(2n-1)/4]λ --- (length as a function of the harmonic)
v=λƒ
v(t)=331.4+0.606T --- (velocity as a function of temperature)
T=1/f --- where T= period

3. The attempt at a solution

Since we can't have negative beat frequencies we know that the second pipe is creating a 30Hz sound.
λ_2=340.39/30=11.35meters (second pipe)
and
λ_1=340.39/10=34.04 meters (first pipe)

We also know that for a closed pipe, the closed end must have a node while the open end must have an antinode. But since we don't know the harmonic, theres not much we can do with that information.

Any ideas?

2. May 23, 2015

### BvU

You missed the bit of info saying the pipes are equal in length ?

3. May 23, 2015

### Ethan_Tab

Im not too sure I understand what you mean.

4. May 23, 2015

### BvU

Well, you have the wavelengths in both pipes and you have the fact that the lengths of the pipes themselves are equal.

5. May 23, 2015

### Ethan_Tab

So if I'm understanding you correctly, you're saying to set the two equation L(n)=[(2n-1)/4]λ_1 and L(n)=[(2n-1)/4]λ_2 equal
[(2n-1)/4]34.04=[(2n-1)/4]11.35
We can solve for the n and in this case
n=0.5

Is that what you mean?

6. May 23, 2015

### BvU

Very close, but no: first of all you need to find an integer number n. In fact you need to find two integer numbers, one for each pipe.
And it becomes easier if for the 34.04 you write $\lambda_1$ and for the other $\lambda_2 = \lambda_1/3$ !

7. May 23, 2015

### BvU

Yet another eureka moment eh ? !

 premature !

Last edited: May 23, 2015
8. May 23, 2015

### Ethan_Tab

Okay so doing that, another system of equation:

λ(2n-1)/4=L
and
λ(2n-1)/12=L

solving for n in equation 1:
n=4L/2λ+1
Putting that into equation 2:
[λ(2(4L/2λ+1)-1]/12
which simplifies into
4L/12=L

Now what?

9. May 23, 2015

### BvU

You need two n. $n_1$ for pipe 1 and $n_2$ for pipe 2. That way
becomes$$\lambda_1 {2n_1-1\over 4} = \lambda_1 {2n_2-1\over 12}$$

10. May 23, 2015

### Ethan_Tab

Im confused as to what I need to solve for.
Also, how would I solve for either n_1 or n_2 given that there is only one equation?

11. May 23, 2015

### BvU

There is one equation. Same as with the other guy, $\lambda_1$ divides out. You are left with one equation and two unknowns. Disaster ? No ! You also know that both are natural numbers.

12. May 23, 2015

### Ethan_Tab

I understand why the two n's must be natural. I don't understand however, how I can solve for either one algebraically given that there is only one equation.

13. May 23, 2015

### BvU

Work out the equation (eliminate the $\lambda_1$ and the factor 4) to its simplest form and see what you get -- in particular, consider which combinations $n_1 , n_2$ satisfy the equation. Then consider what pipe length goes with such solutions.

14. May 23, 2015

### Ethan_Tab

Using guess and check I can see that n_1= 1 when n_2=2

Theres the eureka moment!!!

If we use the λ2n-1/4 with the sets [n_1 and λ_1 or n_2 and λ_2] We get the same (presumably correct) answer of 8.51 meters!

Thanks for the help and patience BvU!

15. May 23, 2015

### BvU

Not done yet: what about higher numbers than 1 and 2 ?

16. May 23, 2015

### Ethan_Tab

Well, n_1=3 and n_2=8 also works

so using these numbers to find the new length I get: 42.55 meters.

17. May 23, 2015

### Ethan_Tab

It appears that there is more then one solution? as there are quite a lot (if not an infinite amount) of possible "n" combinations.

18. May 23, 2015

### BvU

Yes, and in between there is $n_1 =2,\ n_2 = 5$.

My impression is that 8.5 m is already pretty big for a pipe, so you're safe assuming the lowest possible combination is what they want for an answer. But if I were teacher, I'd sure reward the students who point out that there are bigger pipes that also qualify !

In addition, it's already pretty difficult to get such a big pipe to resonate at its second resonance frequency, let alone at its fifth. But that's something the experts should sort out, and I'm just an ordinary physicist...

Well done.

@idc: it's not polite to hijack a thread and interfere this way. Constructive contributions are always welcome, though, if they help the original poster.
--

19. May 23, 2015

### Ethan_Tab

Haha. Again, I appreciate your help immensely.

20. May 24, 2015