Need help, beat frequencies and tuning.

In summary: According to the equation, the wavelength of the flute should be 9.21cm. However, the flute currently has a wavelength of 8.84cm. The difference is .921cm, which is the difference between the intended wavelength and the current wavelength.
  • #1
Numzie
15
0
Need urgent help, beat frequencies and tuning.

Homework Statement


http://img144.imageshack.us/img144/4337/2aee8.th.jpg


Homework Equations





The Attempt at a Solution



You can see my attempt in the picture. I assume that beat frequencies are the absolute value of the difference between the two. The question asks for how far the tube must be extended to the 3rd decimal but i got 9cm which leads me to think its wrong since it has not decimals at all. Can someone please check my math and confirm its correct or lead me in the right way to get it correct. Thanks
 
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  • #2
It seems as if no one could be bothered with problems involving waves and that everyone sticks to the other problems instead. Neither of my two requests for help involving waves have been seen too :(
 
  • #3
Could you explain the formulas for how temperature is involved? I covered waves and stuff last year but we never talked about temperature. I agree that the diagram is correct. The calculation of the wavelength is correct. I'm a little rusty on beat frequencies though..
 
  • #4
Velocity of sound is affected by temperature. [tex]V_Sound{} = 331.5\times.6(Temp\degree{}C[/tex])
 
  • #5
The frequency of the flute as it is now is 266.28 Hz. Since the beat frequency is 3.75Hz and it is too high, the frequency of the tuning fork is 266.28-3.75 Hz = 262.53. This is the frequency you want, so you can relate it to the intended wavelength and length of flute.
 
  • #6
So then 343.5/262.53=1.30842m Wavelength.
1.30842m-1.29m=.01842m
=18.42cm

18.42/2=9.21cm

Same as what i got first time around since frequency can not be negative.
 
  • #7
Why are you subtracting?

In all cases: 1/2 wavelength = length of flute
1/2 (130.842) = length of flute
length of flute = 65.421
Difference = |64.5 - 65.421| = 0.921
 
  • #8
Oh, I see what I did wrong then. I subtracted the total wavelength of both first then divided by 2 instead of using the tube length. Thanks for all the help!
 

FAQ: Need help, beat frequencies and tuning.

1. What is the definition of "beat frequencies"?

Beat frequencies refer to the phenomenon where two sound waves of slightly different frequencies interfere with each other, resulting in a periodic variation in loudness or intensity. This can be observed when two musical notes played on different instruments are slightly out of tune with each other.

2. How can beat frequencies be used to tune musical instruments?

Beat frequencies can be used to tune musical instruments by adjusting the tension or length of the strings until the beats disappear. This indicates that the frequencies of the two notes are now matching and the instruments are in tune.

3. What causes beat frequencies to occur?

Beat frequencies occur when two sound waves with different frequencies interfere with each other. This interference can be constructive, resulting in a louder sound, or destructive, resulting in a quieter sound. The resulting beat frequency is the difference between the two original frequencies.

4. How can beat frequencies be helpful in music production?

Beat frequencies can be used in music production to create interesting and unique sounds. By intentionally tuning instruments slightly out of tune with each other, producers can create beats and patterns that add depth and character to their music.

5. Are beat frequencies only applicable to musical instruments?

No, beat frequencies can occur in any type of sound wave interference, not just musical instruments. They can also be observed in natural phenomena, such as the beating of wings of insects or the pulsing of heartbeats in medical equipment.

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