Flute player initial frequency?

Click For Summary
SUMMARY

The initial frequency of the flute player is determined to be 583 Hz. This conclusion is based on the beat frequency formula, fbeat = |f1 - f2|, where f1 is the frequency of the tuning fork at 587 Hz and fbeat is the observed beat frequency of 4 Hz. By solving the equation, the two possible frequencies are 591 Hz and 583 Hz, with the correct answer being 583 Hz as the flute's lengthening decreases the pitch.

PREREQUISITES
  • Understanding of beat frequency in acoustics
  • Familiarity with the frequency formula fbeat = |f1 - f2|
  • Basic knowledge of sound wave properties
  • Experience with tuning musical instruments
NEXT STEPS
  • Study the principles of beat frequency in sound waves
  • Learn about the physics of musical instrument tuning
  • Explore sound wave visualization techniques using software like Excel
  • Investigate the relationship between flute length and pitch
USEFUL FOR

Musicians, acoustics students, and educators interested in understanding sound wave interactions and tuning principles will benefit from this discussion.

Wimpalot
Messages
35
Reaction score
1

Homework Statement


A flute player hears four beats per second when she compares her note to a 587 Hz tuning fork (the note D). She can match the frequency of the tuning fork by pulling out the "tuning joint" to lengthen her flute slightly. What was her initial frequency?

Homework Equations


Not sure if it actually is relevant but:
fbeat = |f1 - f2|

The Attempt at a Solution


I have no idea what to do for this one. I'm not even entirely sure what is going on
 
Physics news on Phys.org
Have you ever tuned an instrument? I will assume that you have not. If you have two instruments (whatever they are - flutes, guitars, digitally generated sine waves, etc.), and you strike a sound on each of them, if the frequencies of those two sounds are slightly different, you will be able to hear that frequency difference. The effect you will hear is a modulating amplitude. In other words, you will hear the combined sound getting louder and softer at a certain frequency. That frequency is called the beat frequency.

One thing you could do to observe this effect visually is create an Excel sheet. Generate two sine waves - one at, say, 440 Hz (note A4) and another at, say, 444 Hz. Then mathematically add those 2 waveforms and plot the result. You will be able to see the 4 Hz beat frequency.

One other important part of this problem that may not be obvious to everyone is this: When the length of the flute is increased, the pitch (frequency) of the sound decreases.

Edit: Here is a short video clip to demonstrate beat frequency.
 
  • Like
Likes   Reactions: Wimpalot
TomHart said:
Have you ever tuned an instrument? I will assume that you have not. If you have two instruments (whatever they are - flutes, guitars, digitally generated sine waves, etc.), and you strike a sound on each of them, if the frequencies of those two sounds are slightly different, you will be able to hear that frequency difference. The effect you will hear is a modulating amplitude. In other words, you will hear the combined sound getting louder and softer at a certain frequency. That frequency is called the beat frequency.

One thing you could do to observe this effect visually is create an Excel sheet. Generate two sine waves - one at, say, 440 Hz (note A4) and another at, say, 444 Hz. Then mathematically add those 2 waveforms and plot the result. You will be able to see the 4 Hz beat frequency.

One other important part of this problem that may not be obvious to everyone is this: When the length of the flute is increased, the pitch (frequency) of the sound decreases.

Edit: Here is a short video clip to demonstrate beat frequency.


Thanks, that helps me understand what it is but I still have no idea how to do the question
 
Wimpalot said:
fbeat = |f1 - f2|
Your equation is appropriate. If you know fbeat (which you do), and you know f1 (which you do), there are only 2 possible solutions for f2.
 
TomHart said:
Your equation is appropriate. If you know fbeat (which you do), and you know f1 (which you do), there are only 2 possible solutions for f2.
Right of course, thank you
 
Wimpalot said:
Only 2 possible solutions ...
So have you figured out which one?
 
  • Like
Likes   Reactions: Wimpalot
rude man said:
So have you figured out which one?

I think it was 585Hz. Sorry it has been a while since I did this problem. Honestly I think I just guessed at whether it was 585 or 589
 
Wimpalot said:
I think it was 585Hz. Sorry it has been a while since I did this problem. Honestly I think I just guessed at whether it was 585 or 589
Your choices are 587+4 = 591 and 587 - 4 = 583.

Re-consider the problem statement: "She can match the frequency of the tuning fork by pulling out the "tuning joint" to lengthen her flute slightly." (Emphasis mine).
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 43 ·
2
Replies
43
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
5K
Replies
4
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K