- #1

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I just need to help getting started on this one, I don't know how to relate tension to frequency without knowing lamda

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- Thread starter warmfire540
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- #1

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I just need to help getting started on this one, I don't know how to relate tension to frequency without knowing lamda

- #2

dynamicsolo

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The velocity of the waves on the string will be the product of the frequency and wavelength. Changing the tension in the string alters the velocity of the waves, but since the fundamental tone of the string is produced by a standing wave, that wavelength is constant. So you will not need to know it (or the length of the string itself).

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- #3

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The velocity of the waves on the string will be the product of the frequency and wavelength. Changing the tension in the string alters the velocity of the waves, but since the fundamental tone of the string is produced by a standing wave, that wavelength is constant. So you will not need to know it (or the length of the string itself).

Okay, well where would I derive a formula to find the tension, or therefore the change of tension in this case?

- #4

Hurkyl

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You don't have to know lambda in order to manipulate it algebraically.I just need to help getting started on this one, I don't know how to relate tension to frequency without knowing lamda

- #5

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You don't have to know lambda in order to manipulate it algebraically.

How come? I don't get why we don't need lambda? What other equation is there? How do you do it "algebraically?"

- #6

dynamicsolo

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v = f · lam.

For the untuned string, then, you have

v_1 = f_1 · lam_1

and, for the properly tuned string,

v_2 = f_2 · lam_2 .

But the length of the G string doesn't change, so neither does the fundamental standing wavelength at which it vibrates. So we have

lam_1 = lam_2 = lam .

If we now compare the wave velocities, we get

v_2/v_1 = (f_2 · lam)/(f_1 · lam) = f_2/f_1 .

So we don't need to know the length of the string or the fundamental standing wavelength.

You know the relationship between wave velocity, string tension, and string linear mass density, so you can work out the ratio of the tuned versus untuned string tension. (BTW, you don't need to know the linear mass density of the string either: that also will cancel out.)

From the ratio of the tensions, you can find the fractional tension change, and from that, the percentage of tension change in the string.

- #7

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Well from here i see that f2/f1 = 392/380 = 1.0316

So the tuned string's tension is 1.0316 times more than the untuned strings tension.

This is also seen as a 96.94%, or a 3.06% in tension.

- #8

dynamicsolo

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It is true that f2/f1 = v2/v1. But isn't the wave velocity equal to sqrt( T / mu ) ?

- #9

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It is true that f2/f1 = v2/v1. But isn't the wave velocity equal to sqrt( T / mu ) ?

Yeah..so we sub v for the sqrt( T / mu )

so f2/f1= sqrt( T2 / mu )/sqrt( T1 / mu )

1.0316=sqrt( T2 / mu )/sqrt( T1 / mu )

1.0642=(T2/mu) / (T1/mu)

1.0642=T2/T1

So T1*1.0642 = T2

This is a 93.98% difference between strings, where as the difference in percent is 6.02%

- #10

dynamicsolo

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(delta_T / T) x 100% = [ (T2 - T1) / T1 ] x 100% .

The fractional change is equivalent to

(T2 - T1)/T1 = (T2/T1) - (T1/T1) = (T2/T1) - 1 = 1.0642 - 1 = 0.0642 .

So your percentage change should be 6.42% . (Or is 6.02% just a typo?)

- #11

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(delta_T / T) x 100% = [ (T2 - T1) / T1 ] x 100% .

The fractional change is equivalent to

(T2 - T1)/T1 = (T2/T1) - (T1/T1) = (T2/T1) - 1 = 1.0642 - 1 = 0.0642 .

So your percentage change should be 6.42% . (Or is 6.02% just a typo?)

Ah i See! thank you soo much!

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