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

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Homework Help Overview

The problem involves determining the original frequency of Tuning Fork B based on the observed beat frequencies when paired with Tuning Fork A, which has a known frequency of 320 Hz. The context includes the effects of adding elastic bands to Tuning Fork B, which alters its frequency and subsequently the beat frequency.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for beat frequency and how the addition of elastic bands affects the frequency of Tuning Fork B. There is exploration of possible frequencies based on the beat frequency before and after the bands were added.

Discussion Status

Participants are actively engaging in reasoning about the effects of the elastic bands on the frequency of Tuning Fork B. Some have provided calculations and reasoning, while others have questioned assumptions about the setup and the interpretation of results. There is a recognition of the need to clarify the original frequency before the bands were added.

Contextual Notes

There is an emphasis on understanding the relationship between beat frequency and the frequencies of the tuning forks, as well as the impact of external modifications (like elastic bands) on the frequencies involved. Participants are navigating through the implications of their calculations and reasoning.

SpicyWassabi101
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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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