What is the frequency of the string now?

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The discussion revolves around tuning a piano string to the note C at 523 Hz, where the tuner initially hears 2 beats per second, indicating possible string frequencies of 535 Hz or 521 Hz. After tightening the string, the beat frequency increases to 3 beats per second, leading to a calculated frequency of 526 Hz for the string. To determine the percentage change in tension needed to bring the string into tune, a ratio of the frequencies is used, resulting in a tension adjustment of approximately 1.14 percent. The conversation highlights the importance of understanding the relationship between frequency, tension, and the physical properties of the string. Overall, the tuning process involves careful calculations and adjustments based on beat frequencies.
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Homework Statement


While attempting to tune the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
(a) What are the possible frequencies of the string?
(b) When she tightens the string slightly, she hears 3 beats/s What is the frequency of the string now?
(c) By what percentage should the piano tuner now change the tension in the string to bring it into tune?



Homework Equations


I'm not sure what equation to use for part b.
freq= n/2L




The Attempt at a Solution


For part (a) I found that the two possible frequencies are5 23 +/- 2
so 535 Hz or 521 Hz

for part (b) I'm not sure how to get an actual value
From what I understand...when the string is tightened its frequency increase since the beat frequency goes up, its becoming more out of tune.. (is this correct?)

(I think I need part (b) to do (c))

Please help! Thanks!
 
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Hi crazyog,

I think you do part b the same way you did part a. It's just that since they tell you that she tightens the string, that will tell you whether to you plus or minus in your equation. What do you get?
 
Oh ok so I did Freq of beat = absolute value of (F1 - F2)
so...
3 = (524- F2)
solve for F2 = 526 Hz

Then for c I used the equation 2*pi/T = 2*pi*freq
I solve for T for the freq at 526 and at 523...(is that correect)
and I get an answer like T = .00190 and T = .001912 respectively.
not really sure about this...something I am missing ?
 
crazyog said:
Oh ok so I did Freq of beat = absolute value of (F1 - F2)
so...
3 = (524- F2)
solve for F2 = 526 Hz

Then for c I used the equation 2*pi/T = 2*pi*freq
I solve for T for the freq at 526 and at 523...(is that correect)
and I get an answer like T = .00190 and T = .001912 respectively.
not really sure about this...something I am missing ?

I think you might be misreading the equation; this T is the period not the tension. (Remember that tension has units of force.)

What equation relates the tension in the string to other things?
 
Oh ok so I want to use f = v/wavelength which is also equal to (1/2L)(sqrt(T/mu)

can't I set up a ration like F2/F1 = sqrt(T2/T1)
I saw this in the book but I was not sure how they reduced this
then I et T2 = (523/526)^2T1
T2 =0.989T1

but you want to do 1 - 0.989 (right?) so 0.0114
then reduced by 1.14 percent?
 
That looks right to me.

They reduced the fraction to that, because once they formed a ratio the variables that were the same for F1 and F2 cancelled.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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