How to find the tension in a guitar string?

JustinLiang

1. The problem statement, all variables and given/known data
The B-string of a guitar is made of steel (density 7800 kg/m^3), is 63.5 cm long, and has diameter 0.406mm. The fundamental frequency is f = 247.0 Hz. Find the string tension.

2. Relevant equations
F/A = YΔL/L

3. The attempt at a solution
So I know we have the A, the Young's modulus, and the length. But there is no change in length! So there should be no tension/stress. I have no clue where to start, need some hints!
Also how does frequency have anything to do with the tension? Is it just there to throw you off?

Thanks.

turbo

Frequency has a lot to do with tension. The length from the nut to the bridge does not change, but the tension does. You have to adjust the tension in the string using the tuning machine. Higher tension = higher pitch.

Darth Frodo

You should be supplied with the formula..

f = [itex]\frac{1}{2l}[/itex][itex]\sqrt{\frac{T}{\mu}}[/itex]

f = frequency
l = length
T = tension
[itex]\mu[/itex] = mass per unit length

JustinLiang

Quote by Darth Frodo

You should be supplied with the formula..

f = [itex]\frac{1}{2l}[/itex][itex]\sqrt{\frac{T}{\mu}}[/itex]

f = frequency
l = length
T = tension
[itex]\mu[/itex] = mass per unit length

Wow, never seen that before, I think our prof assigned the wrong question. I talked to a lot of my friends and they have no clue what to do with this question either.

Thanks though, now I know :P

robphy

for simple harmonic motion, [itex]F=k x[/itex], where x is the displacement from equilibrium...

Into Newton's Second Law, [itex]m\frac{d^2x}{dt^2}=kx[/itex], you get a differential equation whose solution [itex]x=A\cos(\omega t+\theta_0)[/itex] tells you what
the angular frequency [itex]\omega[/itex] is (and what it depends and doesn't depend on).

Can you use "relevant equation" here?

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turbo Recognitions: Gold Member	Frequency has a lot to do with tension. The length from the nut to the bridge does not change, but the tension does. You have to adjust the tension in the string using the tuning machine. Higher tension = higher pitch.

Darth Frodo Blog Entries: 1 Recognitions: Gold Member	You should be supplied with the formula.. f = [itex]\frac{1}{2l}[/itex][itex]\sqrt{\frac{T}{\mu}}[/itex] f = frequency l = length T = tension [itex]\mu[/itex] = mass per unit length

robphy Blog Entries: 47 Recognitions: Gold Member Homework Help Science Advisor	for simple harmonic motion, [itex]F=k x[/itex], where x is the displacement from equilibrium... Into Newton's Second Law, [itex]m\frac{d^2x}{dt^2}=kx[/itex], you get a differential equation whose solution [itex]x=A\cos(\omega t+\theta_0)[/itex] tells you what the angular frequency [itex]\omega[/itex] is (and what it depends and doesn't depend on). Can you use "relevant equation" here?