# Tension and frequency

by warmfire540
Tags: frequency, tension
 P: 55 The G string of a guitar which should be 392 Hz is playing flat at 380 Hz. What percentage change in the tension of the wire is required so as to tune the string to 392 Hz? I just need to help getting started on this one, I don't know how to relate tension to frequency without knowing lamda
 HW Helper P: 1,662 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).
P: 55
 Quote by dynamicsolo 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?

Emeritus
PF Gold
P: 16,091
Tension and frequency

 Quote by warmfire540 I just need to help getting started on this one, I don't know how to relate tension to frequency without knowing lamda
You don't have to know lambda in order to manipulate it algebraically.
P: 55
 Quote by Hurkyl 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?"
 HW Helper P: 1,662 You have the wave relation 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.
 P: 55 Okay 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.
 HW Helper P: 1,662 It is true that f2/f1 = v2/v1. But isn't the wave velocity equal to sqrt( T / mu ) ?
P: 55
 Quote by dynamicsolo 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%
 HW Helper P: 1,662 That is close, but be careful about the definition. The question asks for the percentage change in the tension of the wire. This will be (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?)
P: 55
 Quote by dynamicsolo That is close, but be careful about the definition. The question asks for the percentage change in the tension of the wire. This will be (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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