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

In summary, the original frequency of Tuning Fork B was 322 Hz. By adding elastic bands, the frequency decreased and the beat frequency dropped to 1 Hz. Testing each possible frequency, it was determined that 321 Hz was the correct frequency, which was obtained by decreasing 322 Hz by 1. Therefore, the original frequency of Tuning Fork B was 322 Hz.
  • #1
SpicyWassabi101
7
0

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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  • #2
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)?
 
  • #3
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?


 
  • #4
Almost! Note that you are looking for the original frequency of tuning fork B (before the bands were added).
 
  • #5
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.
 
  • #6
I agree with your answer and your reasoning.
 
  • #7
Thank you so much for your help! I really appreciate it! :biggrin:
 

What is the concept of beat frequency in relation to tuning forks?

The concept of beat frequency refers to the phenomenon observed when two tuning forks of different frequencies are struck simultaneously. The resulting sound produced is a combination of the two frequencies, creating a pulsating sound known as the beat frequency.

How is the beat frequency calculated using tuning forks?

The formula for beat frequency is f2 = f1 ± fb, where f2 is the frequency of the combined sound, f1 is the frequency of the first tuning fork, and fb is the frequency difference between the two tuning forks. For example, if one tuning fork has a frequency of 500 Hz and the other has a frequency of 505 Hz, the beat frequency would be 5 Hz (500 Hz ± 5 Hz = 505 Hz and 500 Hz – 5 Hz = 495 Hz).

What causes the beat frequency to occur?

The beat frequency occurs due to the interference of sound waves from the two tuning forks. When the two frequencies are close together, the sound waves will either align or cancel each other out at certain points, creating the pulsating sound of the beat frequency.

How can beat frequency be used to tune musical instruments?

By adjusting the frequency of one tuning fork, the beat frequency can be used to tune musical instruments. When the beat frequency is minimized or becomes zero, it means that the two frequencies are in tune with each other. This can be used to tune instruments such as guitars, pianos, and violins.

Can beat frequency be used to determine the accuracy of a tuning fork?

Yes, beat frequency can be used to determine the accuracy of a tuning fork. By comparing the beat frequency with a known frequency, such as that of a standard tuning fork, the accuracy of the tuning fork can be determined. If the beat frequency is significantly different, it may indicate that the tuning fork is not calibrated correctly and may need to be adjusted.

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