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

[dinospamoni]

Homework Statement

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.

Homework Equations

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

$v=f\lambda$

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

 

[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?

 

[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?

 

[haruspex]

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?

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?

 

[Nickie]

I appreciate your attempt at solving this problem. However, there are a few things that need to be clarified and corrected in your solution.

Firstly, the equation you used for wave speed, v=\sqrt{\frac{T}{\mu}}, is only applicable for transverse waves on a string with negligible mass. In this case, the string has a non-negligible mass, so the correct equation to use is v=\sqrt{\frac{T}{\mu}}, where \mu is the linear density (mass per unit length) of the string.

To calculate \mu, you correctly converted the linear mass from grams per meter to kilograms per meter. However, you forgot to include the length of the string in your calculation. The length of the string is given as 1.5 m, so the correct value for \mu is:

\mu=\frac{.0049\frac{kg}{m} \times 1.5 m}{1 m} = 0.00735 \frac{kg}{m}

Using this value for \mu, the correct calculation for wave speed is:

v=\sqrt{\frac{76 N}{0.00735 \frac{kg}{m}}} = 74.33 m/s

Next, for the frequency calculation, you used the wrong equation. The correct equation for the frequency of a standing wave on a string is:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where L is the length of the string. In this problem, L = 1.5 m. Using this equation, we get:

f=\frac{1}{2 \times 1.5 m}\sqrt{\frac{76 N}{0.00735 \frac{kg}{m}}} = 81.18 Hz

Finally, to find the period, you can use the equation T=\frac{1}{f}. Plugging in the calculated frequency, we get:

T=\frac{1}{81.18 Hz} = 0.0123 s

In summary, the correct values for wave speed, frequency, and period are:

Wave speed = 74.33 m/s
Frequency = 81.18 Hz
Period = 0.0123 s

I hope this helps clarify your solution and leads you to the correct answers. Keep up the good work!