# Wave speed, freq, and period of wave on a string

1. May 1, 2013

### dinospamoni

1. The problem statement, all variables and given/known data

A musician frets a guitar string of length 1.5 m at x = 0.34 m with one finger, and simultaneously plucks the string at x = 0.17 m with another finger (raising it to a height h = 2.1 mm. Both fingers are simultaneously removed from the string, and it is allowed to vibrate. The string has a tension of 76 N, and a linear mass m = 4.9 g/m.

2. Relevant equations

$v=\sqrt{\frac{T}{\mu}}$
where
$\mu=\frac{mass\,in\,kg}{Length\,in\,m}$

$v=f\lambda$

3. The attempt at a solution

I started off converting the linear mass:

$4.9\frac{g}{m} \times\frac{1 kg}{1000 g} = .0049\frac{kg}{m}$

Then I plugged that into the equation:

$v=\sqrt{\frac{76 N}{.0049\frac{kg}{m}}}$

and got v=124.54 m/s

From here I used $v=f\lambda$
where $\lambda=.68 m$

and got frequency = 183.147 Hz

Taking the reciprocal of that:

Period = .00546 s

These aren't the right answers, but I don't see where I went wrong

Also, picture attached

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2. May 1, 2013

### haruspex

You don't say exactly what is being asked. You have calculated, correctly I believe, the period as though the fret finger is still in place, but is it what was asked for?

3. May 1, 2013

### dinospamoni

I included what was asked for in the title, but forgot to include it in the question. It's asking for the wave velocity, frequency, and period.

I accounted for the finger being lifted off by making the wavelength twice the interval between the finger and the origin. Is that not right?

4. May 1, 2013

### haruspex

No, that's what it would be if the finger were left in place. But even then, 1.5m is not twice 0.34m. How do you propose to bring 1.5m into the equation?

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