Solving Beat Frequency with Tuning Forks: f2 = f1 ± fb

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SUMMARY

The original frequency of Tuning Fork B is determined to be 322 Hz based on the beat frequency calculations. Initially, a beat frequency of 2 Hz was calculated from 14 beats detected over 7 seconds, leading to possible frequencies of 322 Hz or 318 Hz. After placing elastic bands on Tuning Fork B, the beat frequency dropped to 1 Hz, confirming that the original frequency must be 322 Hz, as the frequency decreased by 1 Hz to yield the final frequency of 321 Hz.

PREREQUISITES
  • Understanding of beat frequency calculations
  • Familiarity with the equation f2 = f1 ± fb
  • Knowledge of frequency measurement in Hertz (Hz)
  • Basic principles of sound wave interference
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  • Study the effects of damping on tuning fork frequencies
  • Explore the physics of sound waves and their interactions
  • Learn about resonance and its impact on tuning forks
  • Investigate the role of tension in vibrating systems
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Students studying physics, particularly those focusing on sound waves and frequency analysis, as well as educators seeking to explain the concept of beat frequencies and their applications.

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Homework Statement


Tuning Fork A has frequency of 320 Hz. When it is sounded with Tuning Fork B, 14 beats are detected in 7.0 seconds. Elastic bands are placed on the tines of Tuning Fork B and the beat frequency drops down to 1.0 Hz. What was the original frequency of Tuning Fork B? Explain your reasoning.

Homework Equations


f2 = f1 ± fb
fb = N / Δt

The Attempt at a Solution


Possible frequencies for Tuning fork B with a beat frequency of 2 Hz:
fb = N/ Δt
fb = 14 / 7.0s = 2.0 Hz

f2 = f1 ± fb
f2 = 320 Hz ± 2.0 Hz = 322 Hz or 318 Hz

Possible frequencies for Tuning Fork B with a beat frequency of 1 Hz:
f2 = 320 Hz ± 1.0 Hz = 321 Hz or 319 Hz

Test each possible frequency:
For 321 Hz:
f2 = 321 - 320 = 1.0 Hz

For 319 Hz:
f2 = 320 - 319 = 1.0 Hz

Both equal 1.0 Hz.

Please help! I have no idea what I did here, I am so confused and this question keeps haunting me :/
Any help would be appreciated!
[/B]
 
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Think about the effect of putting elastic bands on the tines of a tuning fork. Would you expect the rate of vibration of the tines to increase or decrease (compared to without the elastic bands)?
 
The rate of vibration would decrease. So I re-did my answer and it makes a lot more sense than the previous one. Here goes:

fb = N/ Δt
fb = 14 / 7.0s = 2.0 Hz

f2 = f1 ± fb
f2 = 320 Hz ± 2.0 Hz = 322 Hz or 318 Hz

Because of the elastic band on Tuning Fork B, the frequency will decrease. So:
fb= 1.0 Hz
Now each possible frequency is lowered by 1:
322 Hz - 1.0 Hz = 321 Hz
318 Hz - 1.0 Hz = 317 Hz

Now test the two possibilities:
if f2= 317 Hz:
f2= 320 Hz- 317 Hz = 3 Hz ( this is false)

if f2= 321 Hz:
f2= 321 Hz - 320 Hz = 1 Hz ( this is correct because it equals the beat frequency)

So the original frequency of Tuning Fork B is 321 Hz.

Is this correct?


 
Almost! Note that you are looking for the original frequency of tuning fork B (before the bands were added).
 
Ok. So before the bands were added the possible frequencies were 322 Hz and 318 Hz.
So I'm assuming the original frequency would be 322 Hz.

I'm assuming this because 321 Hz was the right frequency for after the bands were added, so since 321 Hz came from decreasing 322 Hz by 1, I would assume that the original frequency would be 322 Hz.

Does that make any sense? correct me if I am wrong please.
 
I agree with your answer and your reasoning.
 
Thank you so much for your help! I really appreciate it! :biggrin:
 

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