Beat Frequency Chart: Solve Problems & Find Original Freq.

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SUMMARY

This discussion focuses on solving beat frequency problems, specifically using a guitar string and a 440 Hz tuning fork. The key findings indicate that when twelve beats are heard in two seconds, the possible frequencies of the string are 434 Hz and 446 Hz, with the original frequency being 434 Hz when the string is tightened. The relationship between tension and frequency is emphasized, noting that tightening the string increases its frequency, which in turn decreases the beat frequency.

PREREQUISITES
  • Understanding of beat frequency concepts
  • Knowledge of frequency and tension relationships in vibrating strings
  • Familiarity with basic physics principles related to sound waves
  • Ability to interpret and create frequency charts
NEXT STEPS
  • Study the principles of beat frequency in sound waves
  • Learn about the effects of tension on the frequency of vibrating strings
  • Explore the mathematical relationships in wave frequency calculations
  • Practice solving beat frequency problems with varying parameters
USEFUL FOR

Students studying physics, particularly those in Grade 11, as well as educators and anyone interested in understanding sound wave interactions and beat frequency phenomena.

Alain12345
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I've posted a few threads on beat frequency, but I still need help. I know that it's really simple, and once I get it I'm going to kick myself for being so stupid, but it just won't get through my head.

I made a chart to help me figure out which possibility is the original one in beat frequency problems. I was only able to complete half of it.

A sample problem is:

When a guitar string is sounded against a 440 Hz tuning fork, twelve beats are heard in two seconds.
a) What are the possible frequencies of the string?
b) If the person tightens the string, the beat frequency decreases. What then is the original frequency of the string, and why?
 

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Alain12345 said:
I've posted a few threads on beat frequency, but I still need help. I know that it's really simple, and once I get it I'm going to kick myself for being so stupid, but it just won't get through my head.

I made a chart to help me figure out which possibility is the original one in beat frequency problems. I was only able to complete half of it.

A sample problem is:

When a guitar string is sounded against a 440 Hz tuning fork, twelve beats are heard in two seconds.
first, do you know how to get the beat frequency from these numbers?
a) What are the possible frequencies of the string?
do you know how to get the two possible frequencies of the string?
b) If the person tightens the string, the beat frequency decreases. What then is the original frequency of the string, and why?
For this part, it's better to clearly break the solution in two steps. First, ignore the tightening part. But I want to first make sure you know how to do the first two parts. Then we will work out through this last part.
 
Yes, I do know how to solve the first two parts. The beat frequency is 6, and the two possibilities are 434 Hz and 446 Hz.
 
Alain12345 said:
Yes, I do know how to solve the first two parts. The beat frequency is 6, and the two possibilities are 434 Hz and 446 Hz.
Ok.

Now here's the tricky part. There are really two steps to the solution and what confuses people comes usually because they are trying to go through both steps at the same time.

For now, ignore completely the fact that beat frequency decreased as opposed to increased. Focus on the fact that the string was tightened. What happens to the frequency of a vibrating string when its tension is increased?
 
The frequency goes up
 
Alain12345 said:
The frequency goes up
Ok. Now is the key point:

If the initial frequency of the string is 634 Hz and its frequency increases, what happens to the beat frequency?

If the initial frequency of the string is 646 Hz and its frequency increases, what happens to the beat frequency?

If you can answer that and look at the question, you will know what the initial frequency of the string was.

What matters in the end is two things: is the string tightened or loosened AND if the beat frequency increases or decreases. For this example (with an initial beat frequency of 6 Hz and a tuning fork of 440 Hz, you should do a chart showing all four cases and giving for each case the initial frequency of the string (which is either 634 or 646 Hz).

For example:

A) If the beat frequency increases as the string is tightened, the initial string frequency was ...

B) If the beta frequency decreases as the string is tightened, the initial string frequency was ...

C) If the beat frequency increases as the string is loosened...

D) if the " " decreases as the string is loosened...


That covers all cases
 
I think I've got it now. But if I were to have to explain it on a test, would this be a good enough answer?

The original frequency is 434 Hz. This is because when the string was tightened, the beat frequency decreased, and the frequency increased. So if you take the two possible frequencies and increase by one, the lesser of the two possibilities (434 Hz) is closer to 440 Hz.
 
Alain12345 said:
I think I've got it now. But if I were to have to explain it on a test, would this be a good enough answer?

The original frequency is 434 Hz. This is because when the string was tightened, the beat frequency decreased, and the frequency increased. So if you take the two possible frequencies and increase by one, the lesser of the two possibilities (434 Hz) is closer to 440 Hz.
You got the right idea although I find the explanation a bit confusing. I would say "when the string was tightened, its frequency increased. Since the beat frequency decreased, we know that the frequency of the string was getting closer to the frequency of the tuning fork, therefore the initial frequency of the string had to be lower than the frequency of the tuning fork, hence 634 Hz"

But that's a bit subjective what is a good verbal explanation (in other words, don't hold me responsible if your prof prefers another wording:biggrin: )

Patrick
 
nrqed said:
You got the right idea although I find the explanation a bit confusing. I would say "when the string was tightened, its frequency increased. Since the beat frequency decreased, we know that the frequency of the string was getting closer to the frequency of the tuning fork, therefore the initial frequency of the string had to be lower than the frequency of the tuning fork, hence 634 Hz"

But that's a bit subjective what is a good verbal explanation (in other words, don't hold me responsible if your prof prefers another wording:biggrin: )

Patrick

lol it's just Grade 11 physics, I don't think my teacher's going to get that picky:wink: . Thanks a lot for your help! :-p
 

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