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I’m not a physics student. Instead I have an electric bass guitar :)

  • I know my strings vibrating-length: 0.800 Meter
  • I know the total weight of this length: 0.016575 Kilo
  • I know my strings resonant frequency: 55.0 Hz
  • I know the wave-velocity: 88.0 Meter per Second

From that, how do I calculate the tension (in Kilo or Newton)?

Ole Sauffaus
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  • Use the equation for tension here to work out the value http://physics.stackexchange.com/questions/202454/modelling-the-sound-wave-of-a-guitar-string-with-an-equation – boyfarrell Nov 03 '15 at 12:46
  • Thanks ... but no thanks ... I have now spent 3 days starring at a multitude of wave-equations, that doesn’t address my precise problem. I’m simply too stupid to grasp them :/ – Ole Sauffaus Nov 03 '15 at 12:51
  • For a fellow bass player: If you know the wave velocity, then $F_T=v^2\mu$ with units of newtons. $\mu$ is the mass per unit length. This is something that you can find in any introductory physics book. – Bill N Nov 03 '15 at 13:11

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The frequency of a standing wave on a guitar string is given by

$$f = \frac{v}{2L}$$

where $v$ is the velocity, and $L$ is the length of the string. It can be shown by using the wave equation (which I'll skip, as it is a more complex derivation) that the velocity of a wave on a string is related to the tension in the string and the mass per unit length, which can be written as:

$$v = \sqrt{\frac{T}{\rho}}$$

where $\rho = m/L$, is the mass per unit length. Combining these, we can write the frequency as:

$$f = \frac{\sqrt{\frac{T}{m/L}}}{2L}$$

Solving for tension, we have:

$$T=4mLf^2$$

tmwilson26
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  • Mmmm... yes ... it is precisely these equations I have been starring at, for 3 days in a row. Could you please show me how my numbers fit into these ... like in a real calculation? – Ole Sauffaus Nov 03 '15 at 13:19
  • @Ole-Sauffaus You just need to put the numbers you gave me into this equation. You have the mass $m = 0.016575 \text{ kg}$, the length $L = 0.8 \text{ m}$, and the frequency $f = 55 \text{ Hz}$. This gives $T = 4(0.016575)(0.8)(55^2) \text{ N} = 160.446 \text{ N}$ – tmwilson26 Nov 03 '15 at 13:23
  • Would that translate to: T = 4 * 0.016575 * 0.8 * pow(2,55); ... Which seems to result in: 1.91097e+15 lol :) – Ole Sauffaus Nov 03 '15 at 13:34
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    @Ole-Sauffaus I'm not sure what you're using to calculate it, so I don't know what pow(2,55) is, but it seems like it would be $2^{55}$, which is why the number is so big. Try pow(55,2). – tmwilson26 Nov 03 '15 at 13:36
  • GOD .... crap ... it is my tool that have wasted 3 days of my time! ... sorry ... My problem is solved. Thanks! – Ole Sauffaus Nov 03 '15 at 13:38